Section Sequences and Continuity in Metric Spaces
We have seen that there are different ways to characterize. For example, there is the \(\epsilon - \delta\) definition and a characterization in terms of neighborhoods. In this section we investigate sequences and limits of sequences in metric spaces, and then provide a characterization of continuous functions in terms of sequences.
Activity 9.2.
A reasonable question to ask is if a limit of a sequence is unique. We will answer that question in this activity. Let \((X,d)\) be a metric space and \((a_n)\) a sequence in \(X\text{.}\) Assume the sequence \((a_n)\) has a limit in \(X\text{.}\) To show that a limit of the sequence \((a_n)\) is unique, we need to show that if \(\lim a_n = a\) and \(\lim a_n = a'\) for some \(a,
a' \in X\text{,}\) then \(a=a'\text{.}\)
Suppose \(\lim a_n = a\) and \(\lim a_n = a'\) for some \(a, a' \in X\text{.}\) Without much to go on it might appear that proving \(a=a'\) is a difficult task. However, if \(d(a,a') \lt \epsilon\) for any \(\epsilon \gt 0\text{,}\) then it will have to be the case that \(a=a'\text{.}\) So let \(\epsilon \gt 0\text{.}\)
(a)
Why must there exist a positive integer \(N\) so that \(d(a_n, a) \lt \frac{\epsilon}{2}\) for all \(n \geq N\text{?}\)
(b)
Why must there exist a positive integer \(N'\) so that \(d(a_n, a') \lt \frac{\epsilon}{2}\) for all \(n \geq N'\text{?}\)
(c)
Now let \(m = \max\{N, N'\}\text{.}\) What can we say about \(d(a_m,a)\) and \(d(a_m,a')\text{?}\) Why?
(d)
Use the triangle inequality to conclude that \(d(a,a') \lt \epsilon\text{.}\) What else can we conclude?
Now we will examine how continuity can be described in terms of sequences. The basic idea is this. Suppose that \(f : \R \to \R\) is continuous at a point \(a\text{.}\) This means that \(f\) has a limit (as a continuous function) at \(a\text{.}\) So if we were to take any sequence \((a_n)\) that converges to \(a\text{,}\) then the continuity of \(f\) implies that \(f(a) = f(\lim a_n) = \lim f(a_n)\text{.}\) That this is both a necessary condition and a sufficient condition for continuity is given in the next theorem.
Theorem 9.7.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, and let \(a \in X\text{.}\) A function \(f:X \to Y\) is continuous at \(a\) if and only if \(\lim f(a_n) = f(a)\) for any sequence \((a_n)\) in \(X\) that converges to \(a\text{.}\)
Proof.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, let \(a \in X\text{,}\) and let \(f: X \to Y\) be a function. Assume that \(f\) is continuous at \(a\text{.}\) We will show that \(\lim f(a_n) = f(a)\) for any sequence \((a_n)\) in \(X\) that converges to \(a\text{.}\) Let \((a_n)\) be a sequence in \(X\) that converges to \(a\) (we know such a sequence exists, namely the sequence \((a)\)). To verify that \(\lim f(a_n) = a\text{,}\) let \(\epsilon \gt 0\text{.}\) The fact that \(f\) is continuous at \(a\) means that there is a \(\delta \gt 0\) so that \(d_Y(f(x),
f(a)) \lt \epsilon\) whenever \(d_X(x,a) \lt \delta\text{.}\) Since \((a_n)\) converges to \(a\text{,}\) we know that there exists a positive integer \(N\) such that \(d_X(a_n, a) \lt \delta\) whenever \(n \geq N\text{.}\) This implies that
\begin{equation*}
d_Y(f(a_n), f(a)) \lt \epsilon \ \text{ whenever } \ n \geq N\text{.}
\end{equation*}
We conclude that if \(f\) is continuous at \(a\text{,}\) then \(\lim f(a_n) = f(a)\) for any sequence \((a_n)\) in \(X\) that converges to \(a\text{.}\)
The proof of the reverse implication is contained in the next activity.
Activity 9.3.
Let
\((X,d_X)\) and
\((Y,d_Y)\) be metric spaces, let
\(a \in X\text{,}\) and let
\(f: X \to Y\) be a function. We prove the remaining implication of
Theorem 9.7, that
\(f\) is continuous at
\(a\) if
\(\lim f(a_n) = f(a)\) for any sequence
\((a_n)\) in
\(X\) that converges to
\(a\text{,}\) in this activity.
(a)
To have an additional assumption with which to work, let us proceed by contradiction and assume that \(f\) is not continuous at \(a\text{.}\) Why can we then say that there is an \(\epsilon \gt 0\) so that there is no \(\delta \gt 0\) with the property that \(d_X(x,a) \lt \delta\) implies \(d_Y(f(x), f(a)) \lt \epsilon\text{?}\)
(b)
To create a contradiction, we will construct a sequence \((a_n)\) that converges to \(a\) while \((f(a_n))\) does not converge to \(f(a)\text{.}\)
(i)
Explain why we can find a positive integer \(K\) such that \(\frac{1}{K} \lt \epsilon\text{.}\)
(ii)
If \(k \gt K\text{,}\) explain why there is an element \(a_k \in B\left(a, \frac{1}{k}\right)\) so that \(d_Y(f(a_k),
f(a)) \geq \epsilon\text{.}\)
(iii)
For \(k \leq K\text{,}\) let \(a_k\) be any element in \(B\left(a, \frac{1}{k}\right)\text{.}\) Explain why \(a\) is a limit of \((a_n)\text{.}\)
(iv)
Explain why \(f(a)\) is not a limit of the sequence \((f(a_n))\text{.}\) What conclusion can we draw, and why?
One way that
Theorem 9.7 is often used is illustrated in the next activity.
Activity 9.4.
Let \(f\) be the function from \(\R\) to \(\R\text{,}\) both with the Euclidean metric, defined by
\begin{equation*}
f(x) = \begin{cases}\sin\left(\frac{1}{x}\right) \amp \text{ if } x \neq 0 \\ 0 \amp \text{ if } x = 0. \end{cases}
\end{equation*}
We consider the \(f\) continuity of \(f\) at \(0\) in this activity.
(a)
Draw a graph of \(f\) on some small interval centered at \(0\text{.}\) Based on the graph, do you think \(f\) has a limit at \(0\text{?}\) Explain. (There is no right answer here, just your intuition based on the graph.)
(b)
At which inputs is \(f(x)=1\text{?}\)
(c)
Use the result of (b) to find a sequence \((a_n)\) that converges to \(0\) for which \(f(a_n) = 1\) for every \(n\text{.}\)
(d)
What does the result of (c) tell us about the continuity of \(f\) at \(0\text{?}\)
While it can sometimes be difficult to prove a fact about all sequences that converge to a point,
Activity 9.4 shows that we can use
Theorem 9.7 to prove that a function
\(f\) is not continuous at an input
\(a\) be finding just one sequence
\((a_n)\) that converges to
\(a\) for which
\(\lim f(a_n) \neq f(a)\text{.}\) We conclude this section with one final note.
IMPORTANT NOTE: Theorem 9.7 tells us that if
\(f : X \to Y\) is a continuous function, then
\(f\) commutes with limits. That is, if
\((a_n)\) is a sequence in
\(X\) that converges to
\(a \in X\text{,}\) then
\begin{equation*}
f(a) = f(\lim a_n) = \lim f(a_n)\text{.}
\end{equation*}
