Definition 1 (Limit of a function) \(\text{Let} \hspace{1.5mm} X \hspace{1.5mm} \text{and} \hspace{1.5mm} Y \hspace{1.5mm} \text{be metric spaces. For a subset} \hspace{1.5mm} E \hspace{1.5mm} \text{of} \hspace{1.5mm} X \hspace{1.5mm} \text{assume there is a function} \hspace{1.5mm} f: E \rightarrow Y. \hspace{1.5mm} \text{Assume} \hspace{1.5mm} p \hspace{1.5mm} \text{is an accumulation} \\ \text{point of} \hspace{1.5mm} E.\)
\(\text{We write either} \hspace{1.5mm} f(x) \to q \hspace{1.5mm} \text{as} \hspace{1.5mm} x \to p, \hspace{1.5mm} \text{or}\)
\[\begin{equation} \large \lim_{x \to p} f(x) = q \end{equation}\]
\(\text{if} \hspace{1.5mm} \underline{\textbf{for every} \hspace{1.5mm} \boldsymbol{\epsilon > 0}}, \hspace{1.5mm} \text{there exists a radius} \hspace{1.5mm} \delta > 0 \hspace{1.5mm} \text{such that for every point} \hspace{1.5mm} x \in \big(N_{\delta}(p) \cap E \big) \setminus \{p\} \hspace{1.5mm} \text{the corresponding image} \\ f(x) \hspace{1.5mm} \text{lies within the neighborhood} \hspace{1.5mm} \underline{\boldsymbol{N_{\epsilon}(q)}}.\)
A consequence of the concept of limit that may be useful is the following. If \(\boldsymbol{f(x)}\) is said to approach \(\boldsymbol{q}\) as \(\boldsymbol{x}\) approaches \(\boldsymbol{p}\) then, for an arbitrary radius \(\boldsymbol{\epsilon > 0}\), it will always be possible to craft a neighborhood \(\boldsymbol{N_{\delta}(p)}\) such that within it there exists a point \(x\) that verifies the following three conditions.
\(x \in E\)
\(x \neq p\)
\(f(x) \in N_{\epsilon}(q)\)
Conditions 1. and 2. stem from \(p\) being an accumulation point of \(E\). Any point \(x \in N_{\delta}(q)\) that fulfills those two conditions automatically satisfies condition 3. —and thus the guaranteed existence of a member of the neighborhood that satisfies the first two actually guarantees fulfillment of all three conditions—.
Theorem 1 (Characterization of the limit) \(\text{Let} \hspace{1.5mm} X, Y, E, f, \hspace{1.5mm} \text{and} \hspace{1.5mm} p \hspace{1.5mm} \text{be as in the Definition above. Then} \hspace{1.5mm} f(x) \to q \hspace{1.5mm} \text{as} \hspace{1.5mm} x \to p \hspace{1.5mm} \text{if and only if}\)
\[\begin{equation} \lim_{n \to \infty} f(s_n) = q \end{equation}\]
\(\text{for every sequence} \hspace{1.5mm} \{s_n\} \hspace{1.5mm} \text{in} \hspace{1.5mm} E \hspace{1.5mm} \text{such that}\)
\[\begin{equation} \lim_{n \to \infty} s_n = p \quad \text{and} \quad \forall n \in \mathbb{N}, \hspace{0.5mm} s_n \neq p. \end{equation}\]
Proof. Because of the definition of convergence, for \(\delta > 0\) all terms of the sequence \(\{s_n\}\) that are indexed by natural numbers greater than or equal to some number \(N\) satisfy \(d(s_n, p) < \delta\). In other words, \(n \geq N \Rightarrow s_n \in N_{\delta}(p)\). Moreover: since the sequence is stated to be in \(E\), those points \(s_n\) also belong to \(E\). And since no term of the sequence is actually equal to \(p\), we conclude \(n \geq N \Rightarrow s_n \in \big(N_{\delta}(p) \cap E \big) \setminus \{p\}\).
Now, as \(\boldsymbol{n}\) approaches infinity, what is \(\boldsymbol{s_n}\) mapped by \(\boldsymbol{f}\) to? Assume \(f(x) \to q\) as \(x \to p\). That is, no matter how narrow we set the neighborhood centered on \(q \in Y\) to be, we are nevertheless always going to be able to find a radius \(\delta > 0\) such that
\[\begin{equation} f\bigg(\big(N_{\delta}(p) \cap E \big) \setminus \{p\}\bigg) \subseteq N_{\epsilon}(q). \end{equation}\]
As \(n\) approaches infinity, we can count on it being big enough so as to satisfy \(\boldsymbol{n \geq N}\), where \(N\) is such that the definition of convergence holds for a radius \(\boldsymbol{\delta}\) which in turn satisfies the condition written above. And therefore, as \(n\) approaches infinity, \(f(s_n) \in N_{\epsilon}(q)\).
Regarding the converse, we will proceed with a proof by contrapositive. Assume \(f(x)\) does not approach \(q\) as \(x\) approaches \(p\). That is
\[\begin{equation} \exists \epsilon > 0, \hspace{0.5mm} \forall \delta > 0, \hspace{0.5mm} \exists x \in \big(N_{\delta}(p) \cap E \big) \setminus \{p\} \hspace{1.5mm} \text{such that} \hspace{1.5mm} f(x) \notin N_{\epsilon}(q). \tag{1} \end{equation}\]
What we intend to conclude is that some sequence \(\{s_n\}\) in \(E\) that converges to \(p\) and whose terms are —all of them— not equal to \(p\) is such that, as \(n\) approaches infinity, there does exist a neighborhood \(N_{\epsilon}(q)\) narrow enough so that \(f(s_n) \notin N_{\epsilon}(q)\).
For every \(n \in \mathbb{N}\), and for a positive \(\epsilon\) that satisfies (1) consider the following set
\[\begin{equation} \large E_n = \bigg\{x \in E \hspace{0.5mm} \bigg| \hspace{0.5mm} x \in N_{\frac{1}{n}}(p) \hspace{1.5mm} \text{and} \hspace{1.5mm} f(x) \notin N_{\epsilon}(q)\bigg\}. \end{equation}\]
It is clear, having assumed (1), that for every \(n \in \mathbb{N}, \hspace{0.5mm} E_n \neq \emptyset\). Now, we will invoke what is known as the axiom of countable choice.
\[\begin{equation} \text{Let} \hspace{1.5mm} \{E_n\} \hspace{1.5mm} \text{be a sequence of non-empty sets. There exists a sequence} \hspace{1.5mm} \{x_n\} \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall n \in \mathbb{N}, \hspace{0.5mm} x_n \in E_n \end{equation}\]
Assuming the axiom of countable choice we claim that there’s an arbitrary sequence \(\{s_n\}\) in \(E\) which, even though it gets closer and closer to \(p\) as \(n\) approaches infinity, does not have a term that gets mapped by \(f\) to the neighborhood \(N_{\epsilon}(q)\). Moreover, since the sequence is arbitrary, we may as well define it so that neither of its terms is equal to \(p\). And with that the proof concludes.
Exercise 1 \(\text{Consider a function} \hspace{1.5mm} f: E \rightarrow \mathbb{R}. \hspace{1.5mm} \text{Let} \hspace{1.5mm} a \hspace{1.5mm} \text{be an accumulation point of} \hspace{1.5mm} E.\)
\(\text{Prove that} \hspace{1.5mm} \large \lim_{x \to a} f(x) = L \Longrightarrow L \in \overline{f(E\setminus \{a\})}.\)
Solution. Essentially, we are asked to prove that if \(f(x)\) approaches \(L\) as \(x\) approaches \(a\) then the point \(L\) either satisfies
\[\begin{equation} \forall r > 0, \hspace{0.5mm} \Big(N_r(L) \cap f(E \setminus \{a\})\Big) \setminus \{L\} \neq \emptyset \tag{2} \end{equation}\]
or it belongs to \(f(E \setminus \{a\}) = \{f(x) \hspace{0.5mm} \big| \hspace{0.5mm} x \in E \hspace{1.5mm} \text{and} \hspace{1.5mm} x \neq a\}\).
For all that we know, there may or may not exist a point \(x \in E\) such that \(f(x) = L\). Thus, we must deal with both cases.
Case 1. \(\boldsymbol{L \notin f(E)}\)
\(L\) does not belong to \(f(E \setminus \{a\})\). Therefore, we must prove that \(L\) is an accumulation point of \(E \setminus \{a\}\). For that to happen (2) must be verified. Well, since \(\boldsymbol{f(x) \to L}\) as \(\boldsymbol{x \to a}\) then for an arbitrary radius \(\boldsymbol{r > 0}\), we are able to craft a neighborhood \(\boldsymbol{N_{\delta}(a)}\) such that within it there will be an \(\boldsymbol{x}\) that verifies \(\underline{\boldsymbol{x \in E}}\), \(\underline{\boldsymbol{x \neq a}}\), and \(\underline{\boldsymbol{f(x) \in N_r(L)}}\).
The underscored expressions stated above unequivocally imply that there exists an element shared between \(f(E\setminus \{a\})\) and \(N_r(L)\) —that is, \(N_r(L) \cap f(E\setminus \{a\}) \neq \emptyset\)—.
Now, in case the intersection of both sets hosts only one point, we need to make sure that that point is not \(L\). Well then, since \(\boldsymbol{L \notin f(E \setminus \{a\})}\), once we intersect this set with the neighborhood \(N_r(L)\) the resulting set —which we already deduced is not empty— will not contain \(\boldsymbol{L}\). And thus, \(L\) is indeed an accumulation point of \(E \setminus \{a\}\).
Case 2. \(\boldsymbol{L \in f(E)}\)
For this case the argument above works up until the point where we conclude \(N_r(L) \cap f(E \setminus \{a\}) \neq \emptyset\). Afterwards, since \(L \in f(E)\), the membership of \(L\) to \(f(E \setminus \{a\})\) is not ruled out, thus making it possible for \(L\) to be the sole element of \(N_r(L) \cap f(E \setminus \{a\})\).
Since nothing we know can lead us to dismiss the membership of \(L\) to \(f(E \setminus \{a\})\) we now have to deal with two cases: \(L = f(a)\) and \(L \neq f(a)\).
Suppose \(L = f(a)\). Then, \(L \notin f(E \setminus \{a\})\). Thus, the same argument used for Case 1 holds and \(L\) is an accumulation point of \(E \setminus \{a\}\).
Suppose \(L \neq f(a)\). Then, \(L \in f(E \setminus \{a\})\).
Either way, it is true that \(L\) either belongs to \(f(E \setminus \{a\})\) or is an accumulation point of that very same set. With that the proof concludes.
Exercise 2 \(\text{Consider a function} \hspace{1.5mm} f: E \rightarrow \mathbb{R} \hspace{1.5mm} \text{and another function} \hspace{1.5mm} g: Y \rightarrow \mathbb{R}. \hspace{1.5mm} \text{Let the range of} \hspace{1.5mm} f \hspace{1.5mm} \text{be a subset of the domain of} \hspace{1.5mm} g. \hspace{1.5mm} \text{Let} \hspace{1.5mm} a \\ \text{be an acccumulation point of} \hspace{1.5mm} E \hspace{1.5mm} \text{and let} \hspace{1.5mm} b \hspace{1.5mm} \text{be —simultaneously— an element of} \hspace{1.5mm} Y \hspace{0.75mm} \textit{and} \hspace{1.5mm} \text{an accumulation point of} \hspace{1.5mm} Y. \hspace{1.5mm} \text{Assume}\)
\[\begin{equation} \large \lim_{x \to a} f(x) = b \quad \text{and} \quad \lim_{y \to b} g(y) = c. \end{equation}\]
\(\text{Prove that if either}\)
\[\begin{equation} c = g(b) \quad \text{or} \quad f^{-1}(b) = \{a\} \end{equation}\]
\(\text{then} \hspace{1.5mm} \large \lim_{x \to a} g(f(x)) = c.\)
Solution. What we are supposed to prove is the following.
\[\begin{equation} \large \forall \epsilon > 0, \hspace{0.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall y \in f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big), \hspace{0.5mm} g(y) \in N_{\epsilon}(c). \end{equation}\]
Since \(f(x) \to b\) as \(x \to a\) then, for an arbitrary radius \(\zeta > 0\), \(f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big) \subseteq N_{\zeta}(b)\).
So far we have deduced that every \(y\) belonging to \(\boldsymbol{f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big)}\) also belongs to \(\boldsymbol{N_{\zeta}(b)}\). Furthermore, it is not hard to note that, since the range of \(\boldsymbol{f}\) is stated to be a subset of the domain of \(\boldsymbol{g}\), every element of \(\boldsymbol{f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big)}\) belongs to \(\boldsymbol{N_{\zeta}(b) \cap Y}\).
Be mindful that, since \(g(y) \to c\) as \(y \to b\) then, for every \(y \in \big(N_{\zeta}(b) \cap Y \big) \setminus \{b\}\), \(g(y) \in N_{\epsilon}(c)\). Therefore, the proof concludes if either we manage to deduce
\[\begin{equation} \boldsymbol{\forall y \in f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big), \hspace{0.5mm} y \neq b \quad \text{or} \quad g(b) \in N_{\epsilon}(c)}. \end{equation}\]
We are essentially asked to verify whether either of these conditions is fulfilled by assuming either of the two proposed assumptions.
Assuming \(\boldsymbol{c = g(b)}\)
In this case we don’t have to worry about some point of \(f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big)\) being equal to \(b\). Such a point will nonetheless be mapped by \(g\) to \(c\) —thus, it will be mapped into \(N_{\epsilon}(c)\)—. Needless to say, with this we have succeeded on this end of the proof.
Assuming \(\boldsymbol{f^{-1}(b) = \{a\}}\)
The only \(x \in E\) that is mapped by \(f\) to \(b\) is \(a\).
Note that for \(y\) to belong to \(f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big)\) the inverse image \(f^{-1}(y)\) is required to be a subset of \(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\) —that is, \(f^{-1}(y)\) must, among other things, not contain the point \(a\)—. So as to make sure that the argument is understood, let us rephrase it in these terms:
\[\begin{equation} \forall y \in f\Big(\big(N_{\delta}(a) \cap E \big) \setminus \{a\}\Big), \hspace{1.5mm} y \neq f(a). \end{equation}\]
Now, since no \(y\) in that set is equal to \(f(a)\), and since \(f(a) = b\) —with \(x \neq a \Rightarrow f(x) \neq b\)— then \(y \neq b\). This, in turn, concludes this end of the proof.
Definition 2 (Continuous function) \(\text{A function} \hspace{1.5mm} f: E \rightarrow Y, \hspace{1.5mm} \text{where} \hspace{1.5mm} Y \hspace{1.5mm} \text{is a metric space and} \hspace{1.5mm} E \hspace{1.5mm} \text{is a subset of a metric space} \hspace{1.5mm} X, \hspace{1.5mm} \text{is said to be continuous at a} \\ \text{point} \hspace{1.5mm} p \in E \hspace{1.5mm} \text{if}\)
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall x \in N_{\delta}(p) \cap E, \hspace{0.5mm} f(x) \in N_{\epsilon}(f(p)). \end{equation}\]
\(\text{If what was stated above is true for every point} \hspace{1.5mm} p \in E \hspace{1.5mm} \text{then} \hspace{1.5mm} f \hspace{1.5mm} \text{is said to be continuous on} \hspace{1.5mm} E.\)
We may express the definition of continuity in terms of the concept of limit. A function \(f: E \rightarrow Y\) is said to be continuous at \(p \in E\) if
\[\begin{equation} \large \lim_{x \to p} f(x) = f(p) \end{equation}\]
Theorem 2 (Characterization of continuity) \(\text{Let} \hspace{1.5mm} X, Y, E, \hspace{1.5mm} \text{and} \hspace{1.5mm} f \hspace{1.5mm} \text{be as in the Definition above, but also let} \hspace{1.5mm} p \hspace{1.5mm} \text{be an accumulation point of} \hspace{1.5mm} E. \hspace{1.5mm} \text{Then} \hspace{1.5mm} f \hspace{1.5mm} \text{is continuous at} \hspace{1.5mm} p \\ \text{if and only if}\)
\[\begin{equation} \lim_{n \to \infty} f(s_n) = f(p) \end{equation}\]
\(\text{for every sequence} \hspace{1.5mm} \{s_n\} \in E \hspace{1.5mm} \text{such that} \hspace{1.5mm} s_n \to p.\)
This proposition greatly resembles that of Theorem 1. Actually, it is easy to see that the proof of this proposition is virtually identical to that of the mentioned Theorem.
Definition 3 (Limit at infinity) \(\text{Let} \hspace{1.5mm} E \subseteq \mathbb{R} \hspace{1.5mm} \text{be a set that is not bounded above in} \hspace{1.5mm} \mathbb{R}. \hspace{1.5mm} \text{Consider a function} \hspace{1.5mm} f: E \rightarrow \mathbb{R}. \hspace{1.5mm} \text{Then we write} \hspace{1.5mm} \lim_{x \to \infty} f(x) = q \hspace{1.5mm} \text{if}\)
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} \exists A > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall x \in \{x \in E \hspace{0.5mm} | \hspace{0.5mm} x > A\}, \hspace{0.5mm} f(x) \in N_{\epsilon}(q). \end{equation}\]
Exercise 3 \(\text{Assume the function} \hspace{1.5mm} f: \mathbb{R} \rightarrow \mathbb{R} \hspace{1.5mm} \text{is continuous and verifies the following condition:}\)
\[\begin{equation} \exists T > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall x \in \mathbb{R}, \hspace{0.5mm} f(x + T) = f(x). \end{equation}\]
\(\text{Show that if} \hspace{1.5mm} f \hspace{1.5mm} \text{is not constant then as} \hspace{1.5mm} x \hspace{1.5mm} \text{approaches infinity there is no} \hspace{1.5mm} q \in \mathbb{R} \hspace{1.5mm} \text{for} \hspace{1.5mm} f(x) \hspace{1.5mm} \text{to approach to.}\)
\(\text{Also show that if} \hspace{1.5mm} f \hspace{1.5mm} \text{is not constant then every element of} \hspace{1.5mm} S = \{T > 0 \hspace{0.5mm} | \hspace{0.5mm} \forall x \in \mathbb{R}, \hspace{0.5mm} f(x + T) = f(x)\} \hspace{1.5mm} \text{is an isolated point of} \hspace{1.5mm} S.\)
Solution. Let us deal with the first proposition through a proof by contrapositive. Assume \(\exists q \in \mathbb{R}\) such that \(\lim_{x \to \infty} f(x) = q\). That is, no matter how narrow we set a neighborhood \(N_{\epsilon}(q)\) to be, we will always be able to find \(A \in \mathbb{R}\) such that \(x > A \Rightarrow f(x) \in N_{\epsilon}(q)\).
Then, for \(p > A\) —taking into account that \(f\) is a periodic function— the following will be satisfied:
\[\begin{equation} f(p + zT) \in N_{\epsilon}(q), \hspace{1.5mm} \text{with} \hspace{1.5mm} z \in \mathbb{Z}. \end{equation}\]
Note that every real number \(x\) can be expressed as the sum of \(zT\) with another real number \(p\) —we just need to adjust the value of \(z\) so that \(p = x - zT\) is greater than \(A\)—. Therefore, \(\boldsymbol{\forall x \in \mathbb{R}, \hspace{1.5mm} f(x) \in N_{\epsilon}(q)}\), with \(\boldsymbol{\epsilon}\) positive and arbitrarily small.
Let us rephrase in these other terms what was stated above: for every \(x \in \mathbb{R}\) it is true that
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} |f(x) - q| < \epsilon. \end{equation}\]
Well then, there is a trusty lemma that should already be quite familiar to us. Let us invoke it once more.
\[\begin{equation} \text{Any two numbers} \hspace{1.5mm} x \hspace{1.5mm} \text{and} \hspace{1.5mm} y \hspace{1.5mm} \text{satisfy} \hspace{1.5mm} x = y \hspace{1.5mm} \text{if and only if} \hspace{1.5mm} \forall \epsilon > 0, \hspace{0.5mm} |x - y| < \epsilon. \tag{3} \end{equation}\]
We conclude, at last, that \(\forall x \in \mathbb{R}, \hspace{1.5mm} f(x) = q\). That is, \(f\) is indeed a constant function —it maps every \(x \in \mathbb{R}\) to \(q\)—.
As for the second proposition, let us proceed again with a proof by contrapositive. Assume at least one element of \(S = \{T > 0 \hspace{0.5mm} | \hspace{0.5mm} \forall x \in \mathbb{R}, \hspace{0.5mm} f(x + T) = f(x)\}\) is an accumulation point of \(S\). That is, \(\exists s \in S\) such that within the neighborhood \(N_{\epsilon}(s)\) —with \(\epsilon\) positive and arbitrary— there exists a point \(s'\) that verifies \(s' \in S\) and \(s' \neq s\). Now, let us assert the following claim:
\[\begin{equation} |zs + z's'| \in S, \hspace{1.5mm} \text{with both} \hspace{1.5mm} z, z' \in \mathbb{Z}. \end{equation}\]
The truth of the claim above is not a matter of careful argumentation but rather a realization that comes from investing some time on thinking about the set \(S\). This set turns out to be closed under integer multiplication and under addition between its own elements. We plug in the absolute value just to make sure that \(|zs + z's'|\) is a positive number —membership of a number \(T\) to \(S\) requires \(T\) to be positive—. In fact, for \(|zs + z's'|\) to belong to \(S\) the following is also required: \(zs \neq -z's'\).
Particularly, by setting \(z = 1\) and \(z' = -1\) we find out that \(\boldsymbol{|s - s'|}\) belongs to \(\boldsymbol{S}\). Also, remember that the distance between \(s\) and \(s'\) is lower than an arbitrarily small positive number \(\epsilon\). Hence, we claim that for every real number \(p\)
\[\begin{equation} f(p + |s - s'|) = f(p) \tag{4} \end{equation}\]
\(\text{and}\)
\[\begin{equation} |p - (p + |s - s'|)| < \epsilon. \tag{5} \end{equation}\]
Consider the set \(V = \Big\{x \in \mathbb{R} \hspace{0.5mm} \Big| \hspace{0.5mm} x = a|s - s'|, \hspace{1.5mm} \text{with} \hspace{1.5mm} a \in \mathbb{Z} \Big\}\). By replacing \(p = 0\) on (4) we obtain \(f(|s - s'|) = f(0)\). Therefore, it is clear that \(V\) is a subset of the inverse image of \(f(0)\) under \(f\). Regarding (5), perhaps the best way to contemplate its consequence is by analogy with the idea of taking a stroll across the real numbers at a pace of \(|s - s'|\) per step. Since that is an extremely small number —infinitesimal, if you will— then the distance between our consecutive footprints —i.e. the distance between consecutive elements of \(V\)— would be minuscule. There would certainly still be gaps —however small they are— between consecutive members of \(V\), though. Therefore, \(V\) fails to comprise the entirety of \(\mathbb{R}\).
Proving that \(f\) is constant is equivalent to proving \(f^{-1}(0) = \mathbb{R}\). Note, on the other hand, that there may be elements of \(f^{-1}(0)\) that are not in \(V\). Now that we have just proven that \(V\) comprises a fair amount of real numbers we have done half the job. Our next task is to show that the gaps in between the elements of \(\boldsymbol{V}\) are as well mapped by \(\boldsymbol{f}\) to \(\boldsymbol{f(0)}\). That is, we need to prove \(\boldsymbol{f^{-1}(0)\setminus V = \mathbb{R}\setminus V}\).
The set \(\mathbb{R} \setminus V\) is equal to \(\large \bigcup_{a \in \mathbb{Z}} \Big(a|s - s'|, \hspace{1mm} (a + 1)|s - s'|\Big)\). It is the union of the gaps in between the elements of \(V\). Notice that for every \(a \in \mathbb{Z}\) the supremum of the corresponding gap is \((a + 1)|s - s'| \in V\). Similarly, the infimum of the corresponding gap is \(a|s - s'| \in V\). That is
\[\begin{equation} \forall a \in \mathbb{Z}, \hspace{0.5mm} \exists p \in V \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall x \in \Big(a|s - s'|, (a + 1)|s - s'|\Big), \hspace{0.5mm} x \in N_{|s - s'|}(p) \end{equation}\]
We are almost done. Now, since \(f\) is stated to be continuous on the entirety of its domain, and since \(p \in V\) belongs to the domain of \(f\) then the funtion is continuous on each of those points \(p\). Allow us to express what we just said in the following terms:
\[\begin{equation}
\forall \zeta > 0, \hspace{0.5mm} \forall x \in \mathbb{R} \setminus V, \hspace{0.5mm} f(x) \in N_{\zeta}(f(0)).
\end{equation}\]
It turns out that every element belonging to the union of gaps in between the members of \(V\) is mapped by \(f\) into an arbitrarily narrow neighborhood around \(f(0)\). Once again, all it takes to prove that those mapped values are in fact exactly equal to \(f(0)\) is the lemma depicted on (3).
The final conclusion is, thus, that \(f\) maps every element of its domain —whether that element lies on \(V\) or on \(\mathbb{R}\setminus V\)— to the exact same value \(f(0)\). That makes \(f\) constant, and this in turn signifies the end of the proof.
Definition 4 (Uniform continuity) \(\text{Let} \hspace{1.5mm} f: E \rightarrow Y. \hspace{1.5mm} \text{Such a function is said to be uniformly continuous if}\)
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \underline{x, \hspace{0.5mm} y \in E \hspace{1.5mm} \text{and} \hspace{1.5mm} |y - x|} < \delta \Longrightarrow |f(y) - f(x)| < \epsilon. \end{equation}\]
Exercise 4 \(\text{Discuss the veracity of the following proposition:}\)
\[\begin{equation} \text{If} \hspace{1.5mm} f \hspace{1.5mm} \text{is continuous on a bounded interval then} \hspace{1.5mm} f \hspace{1.5mm} \text{is uniformly continuous.} \end{equation}\]
Solution. Let’s try a proof by contrapositive. Assume \(f\) is not uniformly continuous. That is
\[\begin{equation} \exists \epsilon > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall \delta > 0, \hspace{1.5mm} \text{some points} \hspace{1mm} x, \hspace{0.5mm} y \in E \hspace{1.5mm} \text{such that} \hspace{1.5mm} |y - x| < \delta \hspace{1.5mm} \text{satisfy} \hspace{1.5mm} |f(x) - f(y)| > \epsilon. \end{equation}\]