4. Invariant set: where dynamics live and remain

A first natural question is whether and when S is the domain of the dynamical system associated with model (1). This amounts asking whether and when S is T-invariant (i.e. Tð ÞS ⊂ S). The complete answer to this question is given by the following proposition and corollary.

In this section, at some point we will consider μ βð Þ as a function of β. So, for consistency, instead of using the simple notation T for the map from model (1), we will use the notation Tμ, <sup>β</sup> which emphasises the explicit dependence of T on the two parameters μ and β.

Proposition 1.5 <sup>T</sup>μ, <sup>β</sup> ð Þ¼ <sup>S</sup> ð Þ <sup>x</sup>, <sup>y</sup> <sup>∈</sup> <sup>þ</sup> � <sup>þ</sup> : <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> ≤ <sup>1</sup> 4 n o:

Remark 1.6 Indeed, we can say more: any point ð Þ u, v ∈ <sup>þ</sup> � <sup>þ</sup> such that

$$\frac{u}{\mu} + \frac{v}{\beta} < \frac{1}{4}$$

admits, exactly, two Tμ, <sup>β</sup> preimages, and they belong to S. Moreover, if ð Þ <sup>u</sup>, <sup>v</sup> <sup>∈</sup> <sup>þ</sup> � <sup>þ</sup> is such that <sup>u</sup> <sup>μ</sup> <sup>þ</sup> <sup>v</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> , then <sup>1</sup> <sup>2</sup> , <sup>2</sup><sup>v</sup> β � � <sup>∈</sup>S is the only <sup>T</sup>μ, <sup>β</sup> preimage of ð Þ u, v :

The line <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> joins the point <sup>μ</sup> <sup>4</sup> , 0 � � with 0, <sup>β</sup> 4 � �: So, when β ≤ 4, it is below the line x þ y ¼ 1 and when β >4 it has points outside S. Consequently, from Proposition 1.5 we get

Corollary 1.7 The simplex S is Tμ, <sup>β</sup> -invariant if and only if β ≤ 4:

Remark 1.8 In fact, it can be easily shown that β ≤4 implies Tμ, <sup>β</sup> ð ÞS ⊈S except when μ ¼ β ¼ 4:

Proof of Proposition 1.5: We start by proving that

$$T\_{\mu,\beta}(\mathbb{S}) \subset \left\{ (\mathfrak{x}, \mathfrak{y}) \in \mathbb{R}^+ \times \mathbb{R}^+ : \frac{\mathfrak{x}}{\mu} + \frac{\mathfrak{y}}{\beta} \le \frac{1}{4} \right\}.$$

Let ð Þ x, y ∈S: We have T xð Þ¼ , y ð Þ μxð Þ 1 � x � y , β xy and,

μxð Þ 1 � x � y , β xy≥ 0 because μ, β >0 and, since ð Þ x, y ∈S, x, y≥0 and x þ y≤1: So, we have proved that T xð Þ , y ∈ <sup>þ</sup> � þ: To end the proof of the above inclusion, we have to show that <sup>μ</sup>xð Þ <sup>1</sup>�x�<sup>y</sup> <sup>μ</sup> <sup>þ</sup> <sup>β</sup> xy <sup>β</sup> ≤ <sup>1</sup> <sup>4</sup> : We have

$$\frac{\mu \varkappa (\mathbf{1} - \varkappa - \jmath)}{\mu} + \frac{\beta \varkappa \jmath}{\beta} = \varkappa (\mathbf{1} - \varkappa - \jmath) + \varkappa \jmath = \varkappa (\mathbf{1} - \varkappa) \le \frac{1}{4} \cdot \varkappa$$

Next we will show that for every ð Þ <sup>u</sup>, <sup>v</sup> <sup>∈</sup> <sup>þ</sup> � <sup>þ</sup> such that <sup>u</sup> <sup>μ</sup> <sup>þ</sup> <sup>v</sup> <sup>β</sup> ≤ <sup>1</sup> 4, there exists ð Þ x, y ∈S such that T xð Þ¼ , y ðμxð Þ 1 � x � y , β xyÞ ¼ ð Þ u, v (i.e. u ¼ μxð Þ 1 � x � y and v ¼ β xy).

If v ¼ 0, it is enough to take y ¼ 0 and x such that μxð Þ¼ 1 � x u: Observe that such point x exists because, in this case,

On Dynamics and Invariant Sets in Predator-Prey Maps DOI: http://dx.doi.org/10.5772/intechopen.89572

$$0 \le u = \mu \left[ \frac{u}{\mu} + \frac{v}{\beta} \right] \le \frac{\mu}{4}.$$

Next we suppose that v> 0: The fact that u∈ <sup>þ</sup> together with <sup>u</sup> <sup>μ</sup> <sup>þ</sup> <sup>v</sup> <sup>β</sup> ≤ <sup>1</sup> <sup>4</sup> implies that 0< v≤ <sup>β</sup> <sup>4</sup> : So, there exist two points 0< y� ≤ <sup>1</sup> <sup>2</sup> ≤ y<sup>þ</sup> < 1 such that

$$
\beta \boldsymbol{\gamma}^- (\mathbf{1} - \boldsymbol{\gamma}^-) = \beta \boldsymbol{\gamma}^+ (\mathbf{1} - \boldsymbol{\gamma}^+) = \boldsymbol{\nu}.
$$

Since <sup>β</sup> <sup>&</sup>gt;0 and 0<sup>&</sup>lt; <sup>y</sup>� <sup>≤</sup> <sup>y</sup><sup>þ</sup> <sup>&</sup>lt; 1, the function z y ð Þ¼ <sup>v</sup> β y from the interval y�, y<sup>þ</sup> ½ � to 1 � yþ, 1 � y� ½ � is a decreasing homeomorphism (observe that we have z y ð Þ; z y� ð Þ¼ <sup>1</sup> � <sup>y</sup>�). Moreover, since <sup>y</sup>� <sup>≤</sup> <sup>1</sup> <sup>2</sup> <sup>≤</sup> <sup>y</sup>þ, we obtain 1 � <sup>y</sup><sup>þ</sup> <sup>≤</sup> <sup>1</sup> <sup>2</sup> ≤1 � y� (see plot above). Consequently,

$$\{\beta \varkappa(\mathbf{1} - \varkappa) : \mathfrak{x} \in [\mathbf{1} - \mathfrak{y}^+, \mathbf{1} - \mathfrak{y}^-] \} = \left[v, \frac{\beta}{4}\right].$$

Hence, there exists a point x ¼ z y ð Þ∈ 1 � yþ, 1 � y� ½ � (of course with y∈ y�, y<sup>þ</sup> ½ �) such that <sup>β</sup> <sup>x</sup>ð Þ¼ <sup>1</sup> � <sup>x</sup> <sup>β</sup> <sup>μ</sup> <sup>u</sup> <sup>þ</sup> <sup>v</sup> because <sup>v</sup><sup>≤</sup> <sup>β</sup> <sup>μ</sup> <sup>u</sup> <sup>þ</sup> <sup>v</sup><sup>≤</sup> <sup>β</sup> <sup>4</sup> : Then, for these particular values of y and x ¼ z y ð Þ, we have β yx ¼ v and

$$
\mu \mathbf{x} (\mathbf{1} - \mathbf{y} - \mathbf{x}) = \mu \mathbf{x} (\mathbf{1} - \mathbf{x}) - \mu \mathbf{y} \mathbf{x} = \frac{\mu}{\beta} (\beta \mathbf{x} (\mathbf{1} - \mathbf{x}) - \beta \mu \mathbf{x}) = \frac{\mu}{\beta} \left( \frac{\beta}{\mu} u + v - v \right) = u.
$$

Next we consider the case β >4: We want to find an invariant subset of S or, equivalently, the domain of definition of T<sup>μ</sup>, <sup>β</sup> as a dynamical system.

We define the one-step escaping set ɛ<sup>μ</sup>, <sup>β</sup> as the set of points z∈S such that T<sup>μ</sup>, <sup>β</sup> ð Þz ∉ S (see Figure 7 for an example). Obviously, ɛ<sup>μ</sup>, <sup>β</sup> ⊂S by definition.

The next proposition gives an estimate of the domain of definition of T<sup>μ</sup>, <sup>β</sup> as a dynamical system (i.e. a T<sup>μ</sup>, <sup>β</sup> -invariant subdomain of S) when β >4 and μ is small enough.

Proposition 1.9 For every <sup>β</sup> <sup>&</sup>gt;4, there exists a unique value <sup>μ</sup><sup>∗</sup> <sup>¼</sup> <sup>μ</sup><sup>∗</sup> ð Þ <sup>β</sup> <sup>∈</sup>ð Þ 0, 4 for which the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μ<sup>∗</sup> <sup>x</sup>ð Þ <sup>1</sup>�<sup>x</sup> <sup>β</sup> �μ<sup>∗</sup> ð Þ<sup>x</sup> and the line <sup>x</sup> <sup>μ</sup><sup>∗</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> intersect at a unique point (see Figure 3). Then, the set Snɛ<sup>μ</sup>, <sup>β</sup> is <sup>T</sup><sup>μ</sup>, <sup>β</sup> -invariant for every <sup>β</sup> <sup>&</sup>gt; 4 and <sup>μ</sup>≤μ<sup>∗</sup> ð Þ <sup>β</sup> :

Proposition 1.9 together with Corollary 1.7 give the splitting of the parameter space according to the shape of the invariant set. Figure 4 and its caption give a graphical description of this splitting together with an account of some dynamical aspects in the different regions (see also Figures 5 and 6).

It is well known that the recurrent dynamics of a dynamical system S, ð Þ T takes place in the non-wandering set of T, <sup>Ω</sup>ð Þ <sup>T</sup> , and <sup>Ω</sup>ð Þ <sup>T</sup> ⊂ ∩<sup>∞</sup> <sup>i</sup>¼0T<sup>i</sup> ð ÞS (see, for instance,

Figure 3.

Three examples of the domain Snɛ<sup>μ</sup>, <sup>β</sup> (in blue) with the one-step escaping set ɛ<sup>μ</sup>, <sup>β</sup> plotted in red for β ¼ 5 and μ ¼ 1:5 (left picture), μ ¼ 2:340246528387⋯ (centre picture), and μ ¼ 3:525 (right picture). The black region shows the set x fð Þ , <sup>y</sup> <sup>∈</sup> <sup>þ</sup> � <sup>þ</sup> : <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> ≤ <sup>1</sup> <sup>4</sup>g⊃T<sup>μ</sup>, <sup>β</sup> ð ÞS ⊃T<sup>μ</sup>, <sup>β</sup> Snɛ<sup>μ</sup>, <sup>β</sup> � �.

#### Figure 4.

The blue region is the one studied by Corollary 1.7: the set S is T<sup>μ</sup>, <sup>β</sup> -invariant. The blue point ( β ¼ μ ¼ 4) is, according to Remark 1.8, the unique point where T<sup>μ</sup>, <sup>β</sup> ð Þ¼ <sup>S</sup> <sup>S</sup>: The red curve is <sup>μ</sup><sup>∗</sup> ð Þ ð Þ <sup>β</sup> , <sup>β</sup> for <sup>β</sup> <sup>∈</sup>ð � <sup>4</sup>, <sup>5</sup> (see Remark 1.10). The green region union with the red curve corresponds to Proposition 1.9: Snɛ<sup>μ</sup>, <sup>β</sup> is T<sup>μ</sup>, <sup>β</sup> invariant. The region at the left of the brown vertical line (i.e. μ<1) corresponds to the parameters for which there exists global convergence to the fixed point P<sup>∗</sup> <sup>1</sup> (Theorem 1.13). The region between the line μ ¼ 1 and the 2 for x∈ ½ � 0, 2 , (

magenta curve ð Þ φ βð Þ, β with φð Þ x ≔ x <sup>x</sup> � <sup>1</sup> for x<sup>∈</sup> ½ � <sup>2</sup>, <sup>5</sup> , corresponds to the parameters for which there

exists global convergence to P<sup>∗</sup> <sup>2</sup> (except for the escaping points and the preimages of P<sup>∗</sup> <sup>1</sup> —Theorem 1.14). The purple dots mark the values of the parameters of the dynamical pictures from Figure 5, and the olive dots mark the values of the parameters of the dynamical pictures from Figure 6.

On Dynamics and Invariant Sets in Predator-Prey Maps DOI: http://dx.doi.org/10.5772/intechopen.89572

#### Figure 5.

Plots of the set ∩<sup>∞</sup> <sup>i</sup>¼<sup>0</sup>T<sup>i</sup> ð ÞS for β ¼ μ ¼ 3:412 (left picture), β ¼ μ ¼ 3:5485 (centre picture), and β ¼ μ ¼ 3:895 (right picture). In Figure 4 we can see the location in the parameter space that corresponds to these three dynamical pictures.

#### Figure 6.

(Top row) The invariant set Snsℛ<sup>μ</sup>, <sup>β</sup> for several values of <sup>β</sup> <sup>&</sup>gt;4 and <sup>μ</sup> <sup>&</sup>gt;μ<sup>∗</sup> ð Þ <sup>β</sup> . In Figure 4 we can see the location in the parameter space that corresponds to these three dynamical pictures. (Bottom row) Escaping regions with the number of iterates needed to go out of the domain represented in a gradient from 1 (black), 2 (dark violet), 3 (light violet) to 50 (yellow) iterates. Note the fractal nature of the invariant set and of the escaping regions (see movie1.mp4 for an animation of the invariant and escaping sets as a function of model parameters).

Lemma 4.1.7 from Ref. [13]). Moreover, both sets <sup>Ω</sup>ð Þ <sup>T</sup> and <sup>∩</sup><sup>∞</sup> <sup>i</sup>¼<sup>0</sup>T<sup>i</sup> ð ÞS are closed and invariant. Then, in the situation of the above proposition (especially in the light of the above remark), we have <sup>Ω</sup>ð Þ <sup>T</sup> ⊂ ∩<sup>∞</sup> <sup>i</sup>¼<sup>0</sup>T<sup>i</sup> ð ÞS ⊈S: To understand the recurrent dynamics of S, ð Þ T , it is clearly interesting (and possible) to characterise the set ∩<sup>∞</sup> <sup>i</sup>¼<sup>0</sup>T<sup>i</sup> ð ÞS (see Figure 9 for some examples for different parameter values).

Of course, as we have already implicitly said, in the region at the left and below the magenta curve (see Figure 4), one only can expect that ∩<sup>∞</sup> <sup>i</sup>¼<sup>0</sup>T<sup>i</sup> ð ÞS will be either P∗ 1 or P<sup>∗</sup> <sup>1</sup> , P<sup>∗</sup> 2 , and, hence, it does not draw much attention.

#### Dynamical Systems Theory

For <sup>β</sup> <sup>&</sup>gt;4 and <sup>μ</sup>>μ<sup>∗</sup> ð Þ <sup>β</sup> , we also want to characterise the invariant set where the dynamics occur. To this end, we define the escaping set R<sup>μ</sup>, <sup>β</sup> as the set of points z∈S such that T<sup>n</sup> <sup>μ</sup>, <sup>β</sup> ð Þz ∉ S for some n ≥1: Clearly,

$$\mathcal{R}\_{\mu,\beta} = \underset{n=0}{\text{U}} \left( \mathbb{S} \cap T\_{\mu,\beta}^{-n} \left( \mathfrak{e}\_{\mu,\beta} \right) \right) = \mathbb{S} \cap \left( \underset{n=0}{\text{U}} \; T\_{\mu,\beta}^{-n} \left( \mathfrak{e}\_{\mu,\beta} \right) \right).$$

As Figure 6 shows, the set SnR<sup>μ</sup>, <sup>β</sup> is (not surprisingly) much more complicated than the sets S and Snɛ<sup>μ</sup>, <sup>β</sup> : This prevents obtaining an analytic characterisation of it, as the one given in Proposition 1.11 for the set Snɛ<sup>μ</sup>, <sup>β</sup> : However, it is always possible (and easy) to obtain numerical approximations to this set for <sup>β</sup> <sup>&</sup>gt;4 and <sup>μ</sup>>μ<sup>∗</sup> ð Þ <sup>β</sup> to gain insight about its shape and topology. Observe (see Figure 6) that the invariant set SnR<sup>μ</sup>, <sup>β</sup> can be fractal.

Remark 1.10. From the proof of Proposition 1.9, it follows that <sup>μ</sup><sup>∗</sup> ð Þ <sup>β</sup> is the unique root in the interval 0, 4 ð Þ of the cubic equation:

$$
\mu^3 + \frac{a\_2(b)}{a\_3(b)}\mu^2 + \frac{a\_1(b)}{a\_3(b)}\mu + \frac{a\_0(b)}{a\_3(b)} = \mathbf{0}
$$

with b ¼ β � 4 and

$$\begin{aligned} a\_3(b) &= b^2\\ a\_2(b) &= -2\left(b^3 + 8b^2 + 16b + 32\right) \\ a\_1(b) &= b^4 + 16b^3 + 96b^2 + 320b + 512, \text{and} \\ a\_0(b) &= -64\left(b^2 + 8b + 16\right). \end{aligned}$$

By means of the Tschirnhaus transformation

$$
\mu = z - \frac{a\_2(b)}{3a\_3(b)} = z + \frac{2}{3b^2} \left( b^3 + 8b^2 + 16b + 32 \right),
$$

the above equation can be transformed into the following equivalent reduced form:

$$z^3 - \frac{p}{3b^4}z + \frac{2q}{27b^6} = 0\tag{9}$$

with

$$\begin{cases} p = -3b^4 \left( \frac{a\_1(b)}{a\_3(b)} - \frac{a\_2(b)^2}{3a\_3(b)^2} \right) \\\\ = b^6 + 16b^5 + 96b^4 + 320b^3 + 1536b^2 + 4096b + 4096, \text{and} \\\ q = \frac{27b^6}{2} \left( \frac{a\_0(b)}{a\_3(b)} - \frac{a\_2(b)a\_1(b)}{3a\_3(b)^2} + \frac{2a\_2(b)^3}{27a\_3(b)^3} \right) \\\\ = b^9 + 24b^8 + 240b^7 + 512b^6 - 3840b^5 - 26112b^4 - 88064b^3 - 245760b^2 - 393216b - 262144, \text{and} \end{cases}$$

Since the linear coefficient of Eq. (9) is negative, it has three real roots, and, by using the trigonometric solution formula for three real root cases, we obtain

On Dynamics and Invariant Sets in Predator-Prey Maps DOI: http://dx.doi.org/10.5772/intechopen.89572

$$\begin{split} z^\* &= 2\sqrt{\frac{1}{3}\frac{p}{3b^4}}\cos\left(\frac{\arccos\left(\left(3\frac{2q}{27b^6}\right)\left(-\frac{1}{2}\frac{3b^4}{p}\right)\sqrt{3\frac{3b^4}{p}}\right)}{3} - \frac{4}{3}\pi\right) \\ &= \frac{2}{3b^2}\sqrt{p}\cos\left(\frac{\pi - \arccos\left(\frac{q}{p\sqrt{p}}\right)}{3} - \frac{4}{3}\pi\right) = -\frac{2}{3b^2}\sqrt{p}\cos\left(\frac{\arccos\left(\frac{q}{p\sqrt{p}}\right)}{3}\right), \end{split}$$

and

$$\mu^\*(\beta) = z^\* - \frac{a\_2(b)}{3a\_3(b)} = \frac{2}{3b^2} \left( -\sqrt{p} \cos\left(\frac{\arccos\left(\frac{q}{p\sqrt{p}}\right)}{3}\right) + b^3 + 8b^2 + 16b + 32} \right).$$

To prove Proposition 1.9, we need a full characterisation of the one-step escaping set when β > 4: This will be obtained in the next proposition.

Proposition 1.11. For every β >4,

$$\mathfrak{c}\_{\mu,\beta} = \left\{ (\mathfrak{x}, \mathfrak{y}) : \left| \mathfrak{x} - \frac{1}{2} \right| < \sqrt{\frac{1}{4} - \frac{1}{\beta}} \text{ and } \frac{1 - \mu \mathfrak{x} (1 - \mathfrak{x})}{(\beta - \mu)\mathfrak{x}} < \mathfrak{y} \le 1 - \mathfrak{x} \right\} \neq \mathfrak{0}$$

(see Figure 7).

Remark 1.12. Observe that ð Þ <sup>x</sup>, <sup>y</sup> <sup>∈</sup>T�<sup>1</sup> <sup>μ</sup>, <sup>β</sup> ð Þ x, y ∈ <sup>þ</sup> � <sup>þ</sup> ð Þ f g : x þ y ¼ 1 if and only if μxð Þþ 1 � x � y β xy ¼ 1 which, in turn, is equivalent to

$$\gamma = \frac{1 - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\beta - \mu)\mathbf{x}} \, . $$

Consequently,

$$\left\{ (\mathbf{x}, \boldsymbol{\uprho}) \in \mathbb{R}^+ \times \mathbb{R}^+ : \boldsymbol{\uprho} = \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \boldsymbol{\uprho})}{(\boldsymbol{\uprho} - \boldsymbol{\uprho}) \mathbf{x}} \right\} = T\_{\mu, \boldsymbol{\uprho}}^{-1} (\{ (\mathbf{x}, \boldsymbol{\uprho}) \in \mathbb{R}^+ \times \mathbb{R}^+ : \boldsymbol{x} + \boldsymbol{\uprho} = \mathbf{1} \})$$

Figure 7.

Two examples of the domain S in blue with the one-step escaping set ɛ<sup>μ</sup>, <sup>β</sup> plotted in red for β ¼ 4:5 and μ ¼ 2 (left picture) and β ¼ 5 and μ ¼ 3:75 (right picture). The one-step escaping set ɛ<sup>μ</sup>, <sup>β</sup> is vertically delimited by the curves <sup>1</sup>�μx 1ð Þ �<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> <sup>&</sup>lt;<sup>1</sup> � x on the interval with endpoints x�<sup>≔</sup> <sup>1</sup> <sup>2</sup> � ffiffiffiffiffiffiffiffiffiffiffi 1 <sup>4</sup> � <sup>1</sup> β q :

and, hence, <sup>ɛ</sup><sup>μ</sup>, <sup>β</sup> is the set of points ð Þ <sup>x</sup>, <sup>y</sup> with <sup>x</sup> � <sup>1</sup> 2 � � � �< ffiffiffiffiffiffiffiffiffiffiffi 1 <sup>4</sup> � <sup>1</sup> β q which are

between the line u þ v ¼ 1 and its T<sup>μ</sup>, <sup>β</sup> preimage (in particular they belong to S). Proof of Proposition 1.11: By assumption we have β >4≥ μ: So, additionally, we have β � μ> 0: We denote

$$\infty^- := \frac{1}{2} - \sqrt{\frac{1}{4} - \frac{1}{\beta}} \quad \text{and} \quad \infty^+ := \frac{1}{2} + \sqrt{\frac{1}{4} - \frac{1}{\beta}}$$

so that <sup>x</sup> � <sup>1</sup> 2 � � � �< ffiffiffiffiffiffiffiffiffiffiffi 1 <sup>4</sup> � <sup>1</sup> β q is equivalent to x∈ x�, x<sup>þ</sup> ð Þ: Thus, since

$$0 < \frac{\sqrt{5} - 1}{2\sqrt{5}} \le \mathfrak{x}^- < \frac{1}{2} < \mathfrak{x}^+ \le \frac{\sqrt{5} + 1}{2\sqrt{5}} < 1,$$

<sup>x</sup> � <sup>1</sup> 2 � � � �< ffiffiffiffiffiffiffiffiffiffiffi 1 <sup>4</sup> � <sup>1</sup> β q implies <sup>x</sup>∈ð Þ 0, 1 : Hence, <sup>μ</sup>xð Þ <sup>1</sup> � <sup>x</sup> <sup>≤</sup>1 and <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> are well defined and non-negative.

To simplify the notation and arguments in the proof, we denote

$$\begin{split} \mathcal{E}\_{\mu,\beta} & \coloneqq \left\{ (\mathbf{x}, \boldsymbol{y}) : \left| \mathbf{x} - \frac{\mathbf{1}}{2} \right| < \sqrt{\frac{1}{4} - \frac{1}{\beta}} \text{ and } \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \boldsymbol{x})}{(\beta - \mu)\mathbf{x}} < \mathbf{y} \le \mathbf{1} - \boldsymbol{x} \right\}, \\ & = \left\{ (\mathbf{x}, \boldsymbol{y}) : \boldsymbol{x} \in (\mathbf{x}^{-}, \boldsymbol{x}^{+}) \text{ and } \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \boldsymbol{x})}{(\beta - \mu)\mathbf{x}} < \mathbf{y} \le \mathbf{1} - \boldsymbol{x} \right\}. \end{split}$$

Then, the proposition states that E<sup>μ</sup>, <sup>β</sup> 6¼ Ø and ɛμ, <sup>β</sup> ¼ E<sup>μ</sup>, <sup>β</sup> : We start by proving that

$$\frac{1 - \mu \mathbf{x} (1 - \mathbf{x})}{(\beta - \mu)\mathbf{x}} < 1 - \mathbf{x} \quad \text{if and only if} \quad \mathbf{x} \in (\mathbf{x}^-, \mathbf{x}^+), \tag{10}$$

which implies that the set E<sup>μ</sup>, <sup>β</sup> is a non-empty subset of S, because <sup>x</sup>�, <sup>x</sup><sup>þ</sup> ð Þ⊂ð Þ 0, 1 and 0<sup>≤</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> . To prove (10) observe that

$$\frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \boldsymbol{\omega})}{(\boldsymbol{\beta} - \boldsymbol{\mu}) \boldsymbol{\omega}} = \mathbf{1} - \boldsymbol{\omega} \Leftrightarrow \frac{\mathbf{1} - \boldsymbol{\beta} \boldsymbol{\omega} (\mathbf{1} - \boldsymbol{\omega})}{(\boldsymbol{\beta} - \boldsymbol{\mu}) \boldsymbol{\omega}} = \mathbf{0} \Leftrightarrow \boldsymbol{\beta} \boldsymbol{\omega} (\mathbf{1} - \boldsymbol{\omega}) = \mathbf{1} \dots$$

On the other hand, x� and x<sup>þ</sup> are the two solutions of the equation <sup>β</sup> <sup>x</sup>ð Þ¼ <sup>1</sup> � <sup>x</sup> <sup>1</sup>: Hence, <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> <sup>¼</sup> <sup>1</sup> � <sup>x</sup> if and only if <sup>x</sup><sup>∈</sup> <sup>x</sup>�, <sup>x</sup><sup>þ</sup> f g: Moreover,

$$\left. \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\boldsymbol{\beta} - \mu)\mathbf{x}} \right|\_{\mathbf{x} = \frac{1}{2}} = \frac{4 - \mu}{2(\boldsymbol{\beta} - \mu)} < \frac{1}{2} = 1 - \frac{1}{2}$$

because β >4: So, (10) holds because <sup>1</sup> <sup>2</sup> ∈ x�, x<sup>þ</sup> ð Þ:

Next we will show that E<sup>μ</sup>, <sup>β</sup> ⊂ɛ<sup>μ</sup>, <sup>β</sup> : For every ð Þ x, y ∈E<sup>μ</sup>, <sup>β</sup> ⊂ S, we have T<sup>μ</sup>, <sup>β</sup> ð Þ¼ x, y ð Þ μxð Þ ð Þ� 1 � y x , β xy with μxð Þ 1 � y � x , β xy ≥0: So,

$$\begin{aligned} \mu \mathbf{x} (\mathbf{1} - \mathbf{y} - \mathbf{x}) + \boldsymbol{\beta} \mathbf{x} \mathbf{y} &= \mu \mathbf{x} (\mathbf{1} - \mathbf{x}) + (\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x} \mathbf{y} > \mu \mathbf{x} (\mathbf{1} - \mathbf{x}) \\ &+ (\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x} \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x}} = \mathbf{1} \end{aligned}$$

Consequently, T<sup>μ</sup>, <sup>β</sup> ð Þ x, y ∉ S, and hence ð Þ x, y ∈ ɛ<sup>μ</sup>, <sup>β</sup> :

On Dynamics and Invariant Sets in Predator-Prey Maps DOI: http://dx.doi.org/10.5772/intechopen.89572

To end the proof of the lemma, we show the other inclusion: ɛ<sup>μ</sup>, <sup>β</sup> ⊂E<sup>μ</sup>, <sup>β</sup> , which is equivalent to SnE<sup>μ</sup>, <sup>β</sup> ⊂Snɛ<sup>μ</sup>, <sup>β</sup> : From above (see again Figure 7) and the fact that for every point ð Þ x, y ∈ S we have μxð Þ 1 � y � x , β xy≥ 0, the inclusion SnE<sup>μ</sup>, <sup>β</sup> ⊂Snɛ<sup>μ</sup>, <sup>β</sup> can be written as

$$\begin{aligned} \{(x,y) : x \in [0,1] \vert (\mathfrak{x}^-, \mathfrak{x}^+) \text{ and } &0 \le y \le 1 - \mathfrak{x} \} \bigcup \\ \{(\mathfrak{x}, y) : x \in (\mathfrak{x}^-, \mathfrak{x}^+) \text{ and } &0 \le y \le \frac{1 - \mu \mathfrak{x} (1 - \mathfrak{x})}{(\beta - \mu)\mathfrak{x}} \} = \mathsf{S} \{ \mathsf{E}\_{\mu, \beta} \subset \mathsf{S} \, \middle| \, \mathfrak{x}\_{\mu, \beta} = \emptyset \} \end{aligned}$$
 
$$\{\underline{x} \in \mathsf{S} : T\_{\mu, \beta}(\underline{x}) \in \mathsf{S} \} = \{ (\mathsf{x}, y) \in \mathsf{S} : \mu \mathbf{x} (1 - y - \mathbf{x}) + \beta \,\mathrm{xy} \le 1 \}.$$

Let us first consider a point ð Þ x, y such that x∈ ½ �n 0, 1 x�, x<sup>þ</sup> ð Þ and y∈ ½ � 0, 1 � x : Since x� and x<sup>þ</sup> are the two solutions of the equation β xð Þ¼ 1 � x 1, it follows that x∈½ �n 0, 1 x�, x<sup>þ</sup> ð Þ is equivalent to β xð Þ 1 � x ≤1: Thus, β > μ gives

$$
\mu \mathbf{x} (\mathbf{1} - \mathbf{y} - \mathbf{x}) + \beta \mathbf{x} \mathbf{y} \le \beta \mathbf{x} (\mathbf{1} - \mathbf{y} - \mathbf{x}) + \beta \mathbf{x} \mathbf{y} = \beta \mathbf{x} (\mathbf{1} - \mathbf{x}) \le \mathbf{1}.
$$

Now we consider a point ð Þ <sup>x</sup>, <sup>y</sup> such that <sup>x</sup><sup>∈</sup> <sup>x</sup>�, <sup>x</sup><sup>þ</sup> ð Þ and 0 <sup>≤</sup>y<sup>≤</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> : In this case, in a similar way as before, we have

$$\begin{aligned} \mu \mathbf{x} (\mathbf{1} - \mathbf{y} - \mathbf{x}) + \boldsymbol{\beta} \mathbf{x} \mathbf{y} &= \mu \mathbf{x} (\mathbf{1} - \mathbf{x}) + (\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x} \mathbf{y} \\ &\leq \mu \mathbf{x} (\mathbf{1} - \mathbf{x}) + (\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x} \frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\boldsymbol{\beta} - \boldsymbol{\mu}) \mathbf{x}} = \mathbf{1}. \end{aligned}$$

Proof of Proposition 1.9: We will use the characterisation of the set ɛμ, <sup>β</sup> given by Proposition 1.11. We start by showing the existence of <sup>μ</sup><sup>∗</sup> <sup>¼</sup> <sup>μ</sup><sup>∗</sup> ð Þ <sup>β</sup> :

Fix <sup>β</sup> <sup>&</sup>gt;4: Clearly, the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> and the line <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> intersect if and only if

$$\frac{1 - \mu \mathbf{x} (1 - \mathbf{x})}{(\beta - \mu)\mathbf{x}} - \left(\frac{\beta}{4} - \mathbf{x}\frac{\beta}{\mu}\right) = \mathbf{0}$$

for some x∈ þ: This equation is equivalent to

$$\frac{(4\mu^2 + 4\beta(\beta - \mu))\mathbf{x}^2 - (4\mu^2 + \beta\mu(\beta - \mu))\mathbf{x} + 4\mu}{4\mu(\beta - \mu)\mathbf{x}} = \mathbf{0}$$

which, in turn, is equivalent to

$$(4\mu^2 + 4\beta(\beta - \mu))\mathfrak{x}^2 - \left(4\mu^2 + \beta\mu(\beta - \mu)\right)\mathfrak{x} + 4\mu = \mathfrak{0}\ldots$$

Thus, the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> and the line <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> intersect at a unique point if and only if the discriminant of the above quadratic equation is zero:

$$\begin{array}{l} \mathbf{0} = \left(4\mu^2 + \beta\,\mu(\beta - \mu)\right)^2 - 16\mu\left(4\mu^2 + 4\,\beta\,(\beta - \mu)\right) \\ = \mu\Big((\beta\,(\beta - 8) + 16)\mu^3 - 2\Big(\beta^2(\beta - 4) + 32\big)\mu^2 + \beta\left(\rho^3 + 64\right)\mu - 64\,\rho^2\Big). \end{array}$$

We need to study the polynomial

$$\begin{aligned} \tilde{P}\_0(\mu) &= (\beta \,(\beta - 8) + 16)\mu^3 - 2 \Big(\beta^2 (\beta - 4) + 32\Big) \mu^2 + \beta \,(\rho^3 + 64)\mu - 64\Big\beta^2 \\ &= a\_3(b)\mu^3 + a\_2(b)\mu^2 + a\_1(b)\mu + a\_0(b), \end{aligned}$$

where the coefficients αið Þ b , with the change of variables β ¼ 4 þ b with b∈ð � 0, 1 , are

$$\begin{aligned} a\_3(b) &:= \beta \left(\beta - 8\right) + 16 = b^2 > 0 \\ a\_2(b) &:= -2\left(\beta^2 \left(\beta - 4\right) + 32\right) = -2\left(b^3 + 8b^2 + 16b + 32\right) < 0 \\ a\_1(b) &:= \beta \left(\beta^3 + 64\right) = b^4 + 16b^3 + 96b^2 + 320b + 512 > 0 \\ a\_0(b) &:= -64\beta^2 = -64\left(b^2 + 8b + 16\right) < 0. \end{aligned}$$

To do it we consider the following sequence of polynomials:

<sup>P</sup>0ð Þ <sup>μ</sup> <sup>≔</sup> <sup>P</sup>~0ð Þ <sup>μ</sup> <sup>α</sup>3ð Þ <sup>b</sup> <sup>¼</sup> <sup>μ</sup><sup>3</sup> <sup>þ</sup> <sup>α</sup>2ð Þ<sup>b</sup> <sup>α</sup>3ð Þ <sup>b</sup> <sup>μ</sup><sup>2</sup> <sup>þ</sup> <sup>α</sup>1ð Þ <sup>b</sup> <sup>α</sup>3ð Þ <sup>b</sup> <sup>μ</sup> <sup>þ</sup> <sup>α</sup>0ð Þ <sup>b</sup> <sup>α</sup>3ð Þ <sup>b</sup> , <sup>P</sup>1ð Þ <sup>μ</sup> <sup>≔</sup> <sup>1</sup> 3 <sup>∂</sup>P0ð Þ<sup>μ</sup> <sup>∂</sup><sup>μ</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>þ</sup> 2α2ð Þ b <sup>3</sup>α3ð Þ <sup>b</sup> <sup>μ</sup> <sup>þ</sup> <sup>α</sup>1ð Þ <sup>b</sup> <sup>3</sup>α3ð Þ <sup>b</sup> , P2ð Þ μ ≔ � 9α3ð Þ b ℜemð Þ P0ð Þ μ , P1ð Þ μ ¼ � <sup>6</sup>α1ð Þ <sup>b</sup> <sup>α</sup>3ð Þ� <sup>b</sup> <sup>2</sup>α2ð Þ <sup>b</sup> <sup>2</sup> <sup>μ</sup> � <sup>9</sup>α0ð Þ <sup>b</sup> <sup>α</sup>3ð Þþ <sup>b</sup> <sup>α</sup>1ð Þ <sup>b</sup> <sup>α</sup>2ð Þ <sup>b</sup> , and P3ð Þ μ ≔P3≔ � ℜemð Þ¼ P1ð Þ μ , P2ð Þ μ � <sup>81</sup>α0ð Þ <sup>b</sup> <sup>2</sup> <sup>α</sup>3ð Þ <sup>b</sup> <sup>2</sup> <sup>þ</sup> <sup>12</sup>α1ð Þ <sup>b</sup> <sup>3</sup> � <sup>54</sup>α0ð Þ <sup>b</sup> <sup>α</sup>1ð Þ <sup>b</sup> <sup>α</sup>2ð Þ <sup>b</sup> <sup>α</sup>3ð Þþ <sup>b</sup> <sup>12</sup>α0ð Þ <sup>b</sup> <sup>α</sup>2ð Þ<sup>b</sup> <sup>3</sup> � <sup>3</sup>α1ð Þ<sup>b</sup> <sup>2</sup> <sup>α</sup>2ð Þ <sup>b</sup> <sup>2</sup> <sup>36</sup>α1ð Þ <sup>b</sup> <sup>2</sup> <sup>α</sup>3ð Þ <sup>b</sup> <sup>2</sup> � <sup>24</sup>α1ð Þ <sup>b</sup> <sup>α</sup>2ð Þ <sup>b</sup> <sup>2</sup> <sup>α</sup>3ð Þþ <sup>b</sup> <sup>4</sup>α2ð Þ <sup>b</sup> <sup>4</sup> <sup>¼</sup> <sup>192</sup>b<sup>10</sup> <sup>þ</sup> <sup>6912</sup>b<sup>9</sup> <sup>þ</sup> <sup>113664</sup>b<sup>8</sup> <sup>þ</sup> <sup>1069056</sup>b<sup>7</sup> <sup>þ</sup> <sup>6438912</sup>b<sup>6</sup> þ <sup>b</sup> <sup>27131904</sup>b<sup>5</sup> <sup>þ</sup> <sup>86507520</sup>b<sup>4</sup> <sup>þ</sup> <sup>214695936</sup>b<sup>3</sup> <sup>þ</sup> <sup>383778816</sup>b<sup>2</sup> <sup>þ</sup> <sup>415236096</sup><sup>b</sup> <sup>þ</sup> <sup>201326592</sup> <sup>b</sup><sup>12</sup> <sup>þ</sup> <sup>32</sup>b<sup>11</sup> <sup>þ</sup> <sup>448</sup>b<sup>10</sup> <sup>þ</sup> <sup>3712</sup>b<sup>9</sup> <sup>þ</sup> <sup>22528</sup>b<sup>8</sup> <sup>þ</sup> <sup>118784</sup>b<sup>7</sup> þ >0 <sup>536576</sup>b<sup>6</sup> <sup>þ</sup> <sup>1900544</sup>b<sup>5</sup> <sup>þ</sup> <sup>5767168</sup>b<sup>4</sup> <sup>þ</sup> <sup>15204352</sup>b<sup>3</sup> <sup>þ</sup> <sup>29360128</sup>b<sup>2</sup> <sup>þ</sup> <sup>33554432</sup><sup>b</sup> <sup>þ</sup> <sup>16777216</sup>

where ℜemð Þ P, Q denotes the remainder of the division of P by Q (i.e. P modulo Q). Since P3ð Þ μ 6¼ 0 for every b, it follows that gcdð Þ¼ P0ð Þ μ , P1ð Þ μ 1, and hence P0ð Þ μ and P1ð Þ μ do not have common roots. In other words, all roots of P0ð Þ μ are simple. Consequently, since <sup>α</sup>3ð Þ <sup>b</sup> <sup>&</sup>gt;0 for every <sup>b</sup>, the equation <sup>P</sup>~0ð Þ¼ <sup>μ</sup> 0 is equivalent to P0ð Þ¼ μ 0, and the above sequence is a Sturm sequence for the polynomial P0ð Þ μ : The following formulae show this Sturm sequence evaluated at μ ¼ 0 and μ ¼ 4, and the signs of these values:

$$\begin{aligned} P\_0(\mathbf{0}) &= \frac{a\_0(b)}{a\_3(b)} < \mathbf{0}, \\ P\_1(\mathbf{0}) &= \frac{a\_1(b)}{3a\_3(b)} > \mathbf{0}, \\ P\_2(\mathbf{0}) &= -9a\_0(b)a\_3(b) + a\_1(b)a\_2(b) = -2\left(b^\top + 24b^6 + 240b^5 + 1088b^4\right) \end{aligned}$$

<sup>þ</sup>2816b<sup>3</sup> <sup>þ</sup> <sup>7680</sup>b<sup>2</sup> <sup>þ</sup> <sup>18432</sup><sup>b</sup> <sup>þ</sup> <sup>16384</sup>Þ<sup>&</sup>lt; 0,

$$\begin{aligned} P\_0(4) &= 64 + 16 \frac{a\_2(b)}{a\_3(b)} + 4 \frac{a\_1(b)}{a\_3(b)} + \frac{a\_0(b)}{a\_3(b)} = \frac{64b^2 + 16a\_2(b) + 4a\_1(b) + a\_0(b)}{b^2} \\ &= \frac{4b^3 + 32b^2 + 128b + 256}{b} > 0, \\\ P\_1(4) &= 16 + 4 \frac{2a\_2(b)}{3a\_3(b)} + \frac{a\_1(b)}{3a\_3(b)} = \frac{48b^2 + 8a\_2(b) + a\_1(b)}{3b^2} = \frac{b^3 + 16b + 64}{3b} > 0, \end{aligned}$$

$$\begin{aligned} P\_2(4) &= -4 \left( 6a\_1(b)a\_3(b) - 2a\_2(b)^2 \right) - 9a\_0(b)a\_3(b) + a\_1(b)a\_2(b) \\ &= -b^2(24a\_1(b) + 9a\_0(b)) + a\_2(b)(a\_1(b) + 8a\_2(b)) \\ &= -2b \Big( b^6 + 20b^5 + 176b^4 + 704b^3 + 1536b(b+1) + 2048 \Big) < 0, \text{and} \\ P\_3(4) &= P\_3 > 0. \end{aligned}$$

So, the sign sequences of the Sturm sequence evaluated at μ ¼ 0 and μ ¼ 4 are f g P0ð Þ 0 , P1ð Þ 0 , P2ð Þ 0 , P3ð Þ 0 ¼ �f g , þ , � , þ which has three changes of sign and f g P0ð Þ 4 , P1ð Þ 4 , P2ð Þ 4 , P3ð Þ 4 ¼ þf g , þ , � , þ which has two changes of sign. Consequently, the polynomial <sup>P</sup>0ð Þ <sup>μ</sup> (and hence the polynomial <sup>P</sup>~0ð Þ <sup>μ</sup> and in turn the above discriminant) has a unique root <sup>μ</sup><sup>∗</sup> <sup>¼</sup> <sup>μ</sup><sup>∗</sup> ð Þ <sup>β</sup> <sup>∈</sup>ð Þ 0, 4 : Moreover, since <sup>P</sup>0ð Þ <sup>0</sup> <sup>&</sup>lt;0 and <sup>P</sup>0ð Þ <sup>4</sup> <sup>&</sup>gt;0, the discriminant is negative for every <sup>μ</sup><sup>∈</sup> 0, <sup>μ</sup><sup>∗</sup> ð Þ and positive for every <sup>μ</sup><sup>∈</sup> <sup>μ</sup><sup>∗</sup> ð � , 4 : This implies that the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> and the line <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> do not intersect whenever <sup>μ</sup><sup>&</sup>lt; <sup>μ</sup><sup>∗</sup> and intersect at a unique point when <sup>μ</sup> <sup>¼</sup> <sup>μ</sup><sup>∗</sup> (see Figure 3). Moreover, for μ small enough and an arbitrary x∈ð � 0, 1 , we have

$$\frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\boldsymbol{\beta} - \mu)\mathbf{x}} - \left(\frac{\boldsymbol{\beta}}{4} - \boldsymbol{\kappa}\frac{\boldsymbol{\beta}}{\mu}\right) > \mathbf{0}.$$

Consequently, since the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> and the line <sup>x</sup> <sup>μ</sup> <sup>þ</sup> <sup>y</sup> <sup>β</sup> <sup>¼</sup> <sup>1</sup> <sup>4</sup> do not intersect for μ<μ<sup>∗</sup> , it follows that

$$\frac{\mathbf{1} - \mu \mathbf{x} (\mathbf{1} - \mathbf{x})}{(\boldsymbol{\beta} - \mu)\mathbf{x}} > \left(\frac{\boldsymbol{\beta}}{4} - \boldsymbol{\pi}\frac{\boldsymbol{\beta}}{\mu}\right) \quad \text{and} \quad \frac{\mathbf{1} - \boldsymbol{\mu}^\* \boldsymbol{\pi} (\mathbf{1} - \boldsymbol{\pi})}{(\boldsymbol{\beta} - \boldsymbol{\mu}^\*)\mathbf{x}} \ge \left(\frac{\boldsymbol{\beta}}{4} - \boldsymbol{\pi}\frac{\boldsymbol{\beta}}{\mu^\*}\right)$$

for every <sup>β</sup> <sup>&</sup>gt;4, <sup>μ</sup><μ<sup>∗</sup> , and <sup>x</sup>∈ð � 0, 1 : On the other hand, by Proposition 1.11, the one-step escaping set <sup>ɛ</sup>μ, <sup>β</sup> is above the parabola <sup>y</sup> <sup>¼</sup> <sup>1</sup>�μxð Þ <sup>1</sup>�<sup>x</sup> ð Þ <sup>β</sup> �<sup>μ</sup> <sup>x</sup> and, by definition, it is contained in S: Consequently, for every β >4 and μ≤μ<sup>∗</sup> ,

$$\left\{ (\mathfrak{x}, \mathfrak{y}) \in \mathbb{R}^+ \times \mathbb{R}^+ : \frac{\mathfrak{x}}{\mu} + \frac{\mathfrak{y}}{\beta} \le \frac{1}{4} \right\} \cap \mathfrak{e}\_{\mu, \beta} = \mathfrak{O},$$

which is equivalent to

$$\mathbf{S} \cap \left\{ (\varkappa, \wp) \in \mathbb{R}^+ \times \mathbb{R}^+ : \frac{\varkappa}{\mu} + \frac{\wp}{\beta} \le \frac{1}{4} \right\} \subset \mathbf{S} \backslash \varepsilon\_{\mu, \beta} \dots$$

On the other hand, for every β > 4≥μ, T<sup>μ</sup>, <sup>β</sup> Snɛ<sup>μ</sup>, <sup>β</sup> ⊂T<sup>μ</sup>, <sup>β</sup> ð Þ<sup>S</sup> and, by definition, T<sup>μ</sup>, <sup>β</sup> Snɛ<sup>μ</sup>, <sup>β</sup> ⊂ S: Then, by Proposition 1.5, for every β > 4 and μ≤μ<sup>∗</sup> ,

$$T\_{\mu,\beta} \left( \mathbb{S} \backslash \varepsilon\_{\mu,\beta} \right) \subset \mathbb{S} \cap T\_{\mu,\beta} \left( \mathbb{S} \right) = \mathbb{S} \cap \left\{ (\infty, y) \in \mathbb{R}^+ \times \mathbb{R}^+ : \frac{\mathbb{x}}{\mu} + \frac{y}{\beta} \le \frac{1}{4} \right\} \subset \mathbb{S} \backslash \varepsilon\_{\mu,\beta} \left( \mathbb{S} \right)$$

This proves that the set Snɛ<sup>μ</sup>, <sup>β</sup> is <sup>T</sup><sup>μ</sup>, <sup>β</sup> -invariant for every <sup>β</sup> <sup>&</sup>gt;4 and <sup>μ</sup>≤μ<sup>∗</sup> ð Þ <sup>β</sup> :
