Definition 1 (Metric space) \(\text{A metric space is a set} \hspace{1.5mm} X \hspace{1.5mm} \text{together with a distance function —also known as metric—} \hspace{1.5mm} d: X \times X \rightarrow \mathbb{R} \hspace{1.5mm} \text{such that for any} \\ \text{elements} \hspace{1.5mm} p, q, r \in X \hspace{1.5mm} \text{the following properties hold:}\)
Particularly, it turns out that \(\boldsymbol{d: (p, q) \mapsto |p - q|}\) satisfies all three conditions. Thus, the set \(\mathbb{R}\) coupled with that distance function is a metric space. From now on we shall consider \(\boldsymbol{d}\) defined the way we have just specified and \(\boldsymbol{X = \mathbb{R}}\).
Definition 2 (Sequence) \(\text{A sequence in a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is a function}\)
\[\begin{align}
f: \mathbb{N} &\rightarrow X \\
n &\mapsto s_n
\end{align}\]
Definition 3 (Convergence) \(\text{If} \hspace{1.5mm} \forall \epsilon > 0, \hspace{0.5mm} \exists N \in \mathbb{N} \hspace{1.5mm} \text{such that} \hspace{1.5mm} n \geq N \Rightarrow d(s_n, s) < \epsilon \hspace{1.5mm} \text{then it is legitimate to assert} \hspace{1.5mm} \lim_{n\to\infty} s_n = s.\)
In plain language: if a sequence \(\{s_n\}\) in a metric space \(X\) is said to converge to \(s\) then, for any real value \(\boldsymbol{\epsilon > 0}\) that we can come up with, it will always be possible to find a natural number \(\boldsymbol{N}\) such that the \(\boldsymbol{N\text{-th}}\) term of the sequence and also each of the terms that come after it lie within a distance less than \(\boldsymbol{\epsilon}\) away from \(s\). Impossibility to find such a natural number \(N\) for some positive \(\epsilon\) is conveyed by the assertion that the sequence does not converge.
Definition 4 (Neighborhood) \(\text{A neighborhood} \hspace{1.5mm} N_r(p) \hspace{1.5mm} \text{of radius} \hspace{1.5mm} r > 0 \hspace{1.5mm} \text{around a point} \hspace{1.5mm} s \hspace{1.5mm} \text{in a metric space} \hspace{1.5mm} X \hspace{1.5mm} \text{is defined as follows:}\)
\[\begin{equation} N_r(p) = \{q \in X | d(p, q) < r\}. \end{equation}\]
If we consider \(X = \mathbb{R}\) then the concept of neighborhood refers to the set \((p - r, p + r)\).
The following is different from the actual definition of convergence —it’s a logically equivalent characterization of that definition—.
Theorem 1 (Characterization of convergence)
\(\text{For any sequence} \hspace{1.5mm} \{s_n\} \hspace{1.5mm} \text{in a metric space} \hspace{1.5mm} X, \hspace{1.5mm} \{s_n\} \hspace{1.5mm} \text{is said to converge to} \hspace{1.5mm} s \in X \hspace{1.5mm} \text{if and only if, for every neighborhood of} \\ s, \hspace{1.5mm} \text{the amount of terms of the sequence that lie outside the neighborhood is finite.}\)
Proof. Let us suppose that for every \(\epsilon > 0\) there is a finite subset of the natural numbers —let’s denote it by \(K\)— whose elements are mapped by the sequence to points \(s_n \in X\) that do not belong to \(N_{\epsilon}(s)\). Note that, for \(\boldsymbol{n \geq maxK + 1}\), each \(\boldsymbol{n}\) is mapped by the sequence to some point \(\boldsymbol{s_n}\) which does belong to \(\boldsymbol{N_{\epsilon}(s)}\). That is, \(N = maxK + 1\) satisfies \(n \geq N \Rightarrow d(s_n, s) < \epsilon\). And with that we have proven the sequence converges to \(s\).
Let us now address the converse. Assume that for every \(\epsilon > 0\) it is possible to find \(N \in \mathbb{N}\) such that each natural number greater than or equal to \(N\) is mapped by the sequence to
Lemma 1
\(\text{Any two numbers} \hspace{1.5mm} x \hspace{1.5mm} \text{and} \hspace{1.5mm} y \hspace{1.5mm} \text{satisfy} \hspace{1.5mm} x = y \hspace{1.5mm} \text{if and only if} \hspace{1.5mm} \forall \epsilon > 0, \hspace{0.5mm} |x - y| < \epsilon\).
Proof. Suppose \(x = y\). Then, \(|x - y| = 0\). Any positive number \(\epsilon\) indeed is greater than \(|x - y|\).
As for the proposition \(\forall \epsilon > 0, \hspace{0.5mm} |x - y| < \epsilon \Longrightarrow x = y\), its demonstration is satisfied through the demonstration of the contrapositive: \(x \neq y \Longrightarrow \exists \epsilon > 0, \hspace{0.5mm} |x - y| \geq \epsilon\).
Suppose \(x \neq y\). Therefore, \(x = z + y\) with \(z \neq 0\). Note that \(|x - y| = |z|\). Now, all we have to do is provide a number \(\epsilon > 0\) such that \(|z| \geq \epsilon\). Note that \(\epsilon = |z|\) is the number we are looking for, and by providing it the proof concludes.
Theorem 2 (Unicity of the limit)
\(\text{For any sequence} \hspace{2mm} \{s_n\} \hspace{2mm} \text{in a metric space} \hspace{2mm} X, \hspace{2mm} \text{if} \hspace{2mm} \{s_n\} \hspace{2mm} \text{converges both to} \hspace{2mm} s \in X \hspace{2mm} \text{and to} \hspace{2mm} s' \in X \hspace{2mm} \text{then} \hspace{2mm} s' = s.\)
Proof. Let’s assume \(\{s_n\}\) does converge to two points —\(s\) and \(s'\), both in \(X\)—. Therefore, the existence of two natural numbers \(N\) and \(N'\) —both dependent on each \(\epsilon > 0\)— such that the following two conditions hold is guaranteed.
\(\exists N \in \mathbb{N} \hspace{2mm} \text{such that} \hspace{2mm} n \geq N \Rightarrow d(s_n, s) < \epsilon\)
\(\exists N' \in \mathbb{N} \hspace{2mm} \text{such that} \hspace{2mm} n \geq N' \Rightarrow d(s_n, s') < \epsilon\)
Note that for \(n \geq max\{N, N'\}\) both inequalities hold. That is, the natural numbers greater than or equal to \(max\{N, N'\}\) are mapped by the sequence to points \(s_n \in X\) that satisfy both \(d(s_n, s) < \epsilon \hspace{1.5mm}\) and \(\hspace{1.5mm} d(s_n, s') < \epsilon\).
Moreover, for \(n \geq max\{N, N'\}\) the following inequality is true: \(d(s_n, s) + d(s_n, s') < 2\epsilon\). We may write the arbitrary positive number \(\epsilon\) in terms of another positive number \(\zeta\) as follows.
\[\begin{equation} \hspace{5mm} \epsilon = \frac{\zeta}{2} \hspace{5mm} \text{with} \hspace{2mm} \zeta > 0 \end{equation}\]
By replacing this on our inequality we obtain \(d(s_n, s) + d(s_n, s') < \zeta\).
Also, since the terms of the sequence lie on a metric space the triangle inequality holds. That is, \(d(s, s') \leq d(s_n, s) + d(s_n, s')\).
Finally, due to transitivity we have the following.
\[\begin{equation} \hspace{5mm} d(s, s') < \zeta \hspace{5mm} \text{with} \hspace{2mm} \zeta > 0 \end{equation}\]
Perhaps it may be useful to state in words what we have so far deduced. For a positive and arbitrary number \(\zeta\), the distance between \(s\) and \(s'\) is less than \(\zeta\). Since \(\zeta\) is arbitrary, we are free to consider it to be as small as we please —so long as it remains positive—. No matter how small we set \(\boldsymbol{\zeta}\) to be, the distance between \(\boldsymbol{s}\) and \(\boldsymbol{s'}\) will always be less than \(\boldsymbol{\zeta}\). Now, because of Lemma 1 we must agree that this implies \(\boldsymbol{s = s'}\).
Definition 5 (Subsequence) \(\text{For any sequence} \hspace{2mm} \{s_n\} \hspace{2mm} \text{in a metric space} \hspace{2mm} X, \hspace{2mm} \text{let} \hspace{2mm} \{n_k\} \hspace{2mm} \text{be a sequence of natural numbers, with} \hspace{2mm} n_1 < n_2 < \ldots \text{Then the} \\ \text{sequence given by} \hspace{2mm} \{s_{n_k}\} \hspace{2mm} \text{is a subsequence of} \hspace{2mm} \{s_n\}.\)
If \(\{s_{n_k}\}\) converges to some \(s \in X\) then \(s\) is a subsequential limit of \(\{s_n\}\).
Lemma 2
\(\text{If the sequence} \hspace{2mm} g: \mathbb{N} \rightarrow \mathbb{N}, \hspace{1.5mm} g: k \mapsto n_k \hspace{2mm} \text{satisfies} \hspace{2mm} n_1 < n_2 < \ldots, \hspace{2mm} \text{then} \hspace{2mm} \forall k \in \mathbb{N}, \hspace{0.5mm} k \leq n_k.\)
Proof. Our chosen procedure to prove this lemma is induction.
Base case: \(g(1) = n_1\) must be a natural number. Since no natural number is less than the number one then \(1 \leq n_1\).
Inductive hypothesis: \(k - 1 \leq n_{k - 1}\).
Inductive step: assume \(n_{k - 1} < n_k\). By incorporating the inductive hypothesis we have \(k - 1 \leq n_{k - 1} < n_k\). Moreover, due to transitivity we have \(k - 1 < n_k\). Since we are dealing with natural numbers this last expression can also be written as \(k \leq n_k\). With that we conclude the proof.
Theorem 3 (Convergence of all subsequences)
\(\text{For any sequence} \hspace{2mm} \{s_n\} \hspace{2mm} \text{in a metric space} \hspace{2mm} X, \hspace{2mm} \{s_n\} \hspace{2mm} \text{converges to} \hspace{2mm} s \in X \hspace{2mm} \text{if and only if every subsequence of} \hspace{2mm} \{s_n\} \hspace{2mm} \text{converges} \\ to \hspace{2mm} s.\)
Proof. Suppose \(\{s_k\}\) converges to \(s\). That is, suppose for every \(\epsilon > 0\) it is possible to provide a natural number \(N\) such that \(k \geq N \Longrightarrow d(s_k, s) < \epsilon\).
The terms of the sequence are originally obtained by evaluating a function \(f: k \mapsto s_k\) on each \(k \in \mathbb{N}\). If we evaluate \(f\) on the elements \(n_k\) of the range of a function \(g: \mathbb{N} \rightarrow \mathbb{N}\) instead we will get for each \(n_k\) a corresponding \(f \circ g(k) = s_{n_k}\).
Consider \(E = \{k \in \mathbb{N} | k \geq N\}\) —such a set \(E\) is crafted for each \(\epsilon > 0\), with a corresponding value of \(N\) that verifies the definition of convergence—. The question we are addressing is whether membership of \(k\) to \(E\) guarantees \(d(s_{n_k}, s) < \epsilon\).
Now, the function \(g\) is not just any function. It satisfies \(n_1 < n_2 < \ldots .\) Then —this we know because of Lemma 2— for every \(k \in \mathbb{N}, \hspace{0.5mm} k \leq n_k\). Particularly, since \(E \subseteq \mathbb{N}\), each \(k \in E\) is mapped by \(g\) to a corresponding \(n_k\) that is greater than or equal to \(k\) itself. From here it all boils down to a matter of transitivity: since \(n_k \geq k\) and \(k \geq N\) then \(n_k \geq N\). That is, if \(k\) belongs to \(E\) then \(n_k\) belongs to \(E\) as well —and each of these \(n_k\) are mapped by \(f\) to points \(s_{n_k}\) that do verify \(d(s_{n_k}, s) < \epsilon\)—.
We must show now that if every subsequence of \(\{s_n\}\) converges to \(s\) the original sequence does also converge to \(s\). Note that the function \(g: k \mapsto n_k = k\) is such that \(f \circ g(k) = s_k\). Also note that the function we just defined satisfies \(n_1 < n_2 < \ldots .\) That is, \(\{s_k\}\) turns out to be a subsequence of itself. Now, since all subsequences converge to \(s\), this particular subsequence does too.
Definition 6 (Bounded set) \(\text{The subset} \hspace{2mm} E \hspace{2mm} \text{of a metric space} \hspace{2mm} X \hspace{2mm} \text{is bounded if} \hspace{2mm} \exists q \in X, \hspace{0.5mm} \exists M \in \mathbb{R} \hspace{2mm} \text{such that} \hspace{2mm} \forall p \in E, \hspace{2mm} d(p, q) \leq M.\)
In plain, everyday language: any subset of some metric space is said to be bounded if we are able to find some point \(q\) in the metric space such that every element of the subset is at most at a distance \(M\) from \(q\).
Note that in order to show that a set is bounded it suffices to provide one value of \(q\). Also note that the choice of \(q\) conditions the value of \(M\).
Theorem 4 (Convergence implies boundedness)
\(\text{For any sequence} \hspace{2mm} \{s_n\} \hspace{2mm} \text{in a metric space} \hspace{2mm} X, \hspace{2mm} \text{if} \hspace{2mm} \{s_n\} \hspace{2mm} \text{converges then the set} \hspace{2mm} f(\mathbb{N}) \subset X \hspace{2mm} \text{is bounded}\).
Proof. Assume \(\{s_n\}\) converges to a point \(s \in X\). Consider a fixed positive number \(\epsilon^{\ast}\). Any term of the sequence lies either within the neighborhood \(N_{\epsilon^{\ast}}(s)\) or outside of it. That is
\[\begin{equation} f(\mathbb{N}) = \Big(f(\mathbb{N}) \cap N_{\epsilon^{\ast}}(s)\Big) \cup \Big(f(\mathbb{N}) \setminus N_{\epsilon^{\ast}}(s)\Big) \end{equation}\]
For all those \(s_n\) within the neighborhood the inequality \(d(s_n, s) \leq \epsilon^{\ast}\) is satisfied. Regarding the points \(s_n\) that do not belong to the neighborhood, there only exists a finite amount of these —the reason we are so sure about that is Theorem 1—. In case the set \(f(\mathbb{N}) \setminus N_{\epsilon^{\ast}}(s)\) is empty then we can provide \(M = \epsilon^{\ast}\) as evidence of the fact that \(\{s_n\}\) is bounded. In case \(f(\mathbb{N}) \setminus N_{\epsilon^{\ast}}(s)\) is not empty then the farthest point \(s_n\) from \(s\) must lie on that set —and therefore \(M\) may be a number greater than \(\epsilon^{\ast}\)—.
The terms of the sequence that lie on \(f(\mathbb{N}) \setminus N_{\epsilon^{\ast}}(s)\) are those that correspond to \(n < N\), where \(N\) is the lowest natural number that verifies \(n \geq N \Rightarrow d(s_n, s) < \epsilon^{\ast}\). That is, the first \(N - 1\) terms of the sequence do not belong to the neighborhood \(N_{\epsilon^{\ast}}(s)\).
The value \(M\) we shall provide is either (in case the neighborhood has a radius \(\epsilon^{\ast}\) wide enough for all terms of the sequence to be contained in it) \(\hspace{0.5mm} \epsilon^{\ast}\) or (in case the neighborhood does not comprise the entirety of the sequence) the distance between \(s\) and the farthest \(s_n\) from it. All this is condensed into the following:
\[\begin{equation} M = max\{d(s_1, s), d(s_2, s), \ldots, d(s_{N - 1}, s), \epsilon^{\ast}\} \end{equation}\]
And that wraps it all up. A sequence that converges to \(s\) is bounded because we can provide as evidence \(q = s\) and \(M\) defined as above.
Definition 7 (Monotonic sequence) \(\text{For any sequence} \hspace{2mm} \{s_n\} \hspace{2mm} \text{in an ordered field} \hspace{2mm} F, \hspace{2mm} \{s_n\} \hspace{2mm} \text{is monotonically increasing if} \hspace{2mm} s_n \leq s_{n + 1} \hspace{2mm} \text{for every} \hspace{2mm} n \in \mathbb{N}.\)
\(\text{Analogously,} \hspace{2mm} \{s_n\} \hspace{2mm} \text{is monotonically decreasing if} \hspace{2mm} s_n \geq s_{n + 1} \hspace{2mm} \text{for every} \hspace{2mm} n \in \mathbb{N}.\)
Theorem 5 (Bounded monotonic sequences)
\(\text{Let} \hspace{2mm} \{s_n\} \hspace{2mm} \text{be a monotonic sequence in an ordered field} \hspace{2mm} F, \hspace{2mm} \text{and let} \hspace{2mm} F \hspace{2mm} \text{have the least upper bound property. Then} \hspace{2mm} \{s_n\} \\ \text{converges in} \hspace{2mm} F \hspace{2mm} \text{if and only if} \hspace{2mm} \{s_n\} \hspace{2mm} \text{is bounded}.\)Theorem 6 (Bolzano-Weirstrass theorem)
Solution. Note that the lowest natural number \(N\) that satisfies \(n \geq N \Rightarrow d(s_n, s) < \epsilon\) for an arbitrary \(\epsilon > 0\) is \(N = 1\).
How can we be so sure about it? Suppose, instead, that the lowest natural number \(N\) that satisfies the definition of convergence is greater than the number one. Note that, since \(\mathbb{N} \subset \mathbb{R}, \hspace{0.5mm} N\) is a real number. The same can be argued for \(p\). Now, no matter how big \(N\) is, we will always be able to find some natural number \(m\) such that \(mp > N\). This is true because of the Archimedean property of the real numbers. Moreover: because of transitivity, if \(mp\) is greater than \(N\), so is \(mp + 1\).
On the other hand, since \(\{s_n\}\) is periodic, the function \(f: n \mapsto s_n\) is such that, for every \(s_n\) that belongs to the range \(f(\mathbb{N})\), the set of all elements of \(\mathbb{N}\) whose image under \(f\) is \(s_n\) is \(f^{-1}(s_n) = \{n, \hspace{0.5mm} n + p, \hspace{0.5mm} n + 2p, \hspace{0.5mm} \ldots \}.\) Let us define that set by a predicate rather than by the enumeration of its elements.
\[\begin{equation} f^{-1}(s_n) = \Big\{mp + n \hspace{0.5mm} \bigg| \hspace{2mm} m \in \mathbb{N} \cup \{0\}\Big\} \end{equation}\]
Then, \(mp + 1 > N\) not only implies \(d(s_{mp + 1}, s) < \epsilon\) but also \(d(s_1, s) < \epsilon\). Similarly, \(mp + 2 > N\) implies \(d(s_2, s) < \epsilon\). Generalizing: since both \(\hspace{1mm} \boldsymbol{mp + n} \hspace{1mm}\) and \(\hspace{1mm} \boldsymbol{n} \hspace{1mm}\) are mapped to the very same number, \(\hspace{1mm} \boldsymbol{mp + n > N} \hspace{1mm}\) implies \(\hspace{1mm} \boldsymbol{d(s_n, s) < \epsilon}\). And since —again, because of the Archimedean property— the inequality \(mp + n > N\) is verified for every \(n \in \mathbb{N}\), we have been succesful in proving the following assertion.
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} \forall n \in \mathbb{N}, \hspace{0.5mm} d(s_n, s) < \epsilon \end{equation}\]
So far so good. Note that the range \(f(\mathbb{N})\) comprises only \(p\) elements. That is, —from what we know— the sequence only exhibits \(p\) distinct values: \(s_1, \hspace{0.5mm} \ldots, \hspace{0.5mm} s_p\). For \(n = 1, \hspace{0.5mm} \ldots, \hspace{0.5mm} p\) the following is verified.
\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} |s_n - s| < \epsilon \end{equation}\]
Because of Lemma 1, we know this implies \(s_n = s\) for each of the first \(p\) natural numbers. That is, the sequence actually exhibits just one distinct value. Therefore, we prove at last that \(s_1 = s_2 = \ldots.\)
Exercise 2 (Convergence of odd and even subsequences to the same limit) \(\text{Suppose there are two sequences} \hspace{2mm} \{x_n\}, \hspace{2mm} \{y_n\} \hspace{2mm} \text{such that} \hspace{1.5mm} \lim x_n = \lim y_n = a. \hspace{2mm} \text{From these sequences we define} \hspace{2mm} \{z_n\} \hspace{2mm} \text{as follows.}\)
\[\begin{equation} \large z_{2n - 1} = x_n \quad \normalsize \text{and} \large \quad z_{2n} = y_n \end{equation}\]
\(\text{Prove that} \hspace{1.5mm} \lim z_n = a.\)
Solution. What we are going to do is prove the contrapositive of this proposition. That is, we will prove that if \(\{z_n\}\) does not converge to \(a\) then one of the sequences \(\{x_n\}\) and \(\{y_n\}\) does not converge, or perhaps neither does.
To assume \(\lim z_n \neq a\) is to assume
\[\begin{equation} \exists \end{equation}\]