Section 1.4 Continuity
The following definition should be compared with the corresponding concept for functions on an interval, say.
Definition 1.42.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A map \(f \maps X
\to Y\) is continuous at a point \(x_0 \in X\) if
\begin{gather*}
\forall \varepsilon \gt 0
\,
\exists \delta \gt 0
\st
\forall x \in X,\\
d_X(x,x_0) \lt \delta
\implies
d_Y(f(x),f(x_0)) \lt \varepsilon\text{.}
\end{gather*}
If \(f\) is continuous at every point of \(X\text{,}\) then we say that \(f\) is continuous.
There are also stronger versions of continuity.
Definition 1.43.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A map \(f \maps X
\to Y\) is called uniformly continuous if
\begin{gather*}
\forall \varepsilon \gt 0
\,
\exists \delta \gt 0
\st
\forall x_1, x_2 \in X,\\
d_X(x_1, x_2) \lt \delta
\implies
d_Y(f(x_1), f(x_2)) \lt \varepsilon\text{.}
\end{gather*}
Definition 1.44.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A map \(f \maps X
\to Y\) is called Lipschitz continuous if there exists a number \(L \ge 0\) such that
\begin{equation*}
d_Y(f(x_1), f(x_2)) \le L d_X(x_1, x_2)
\end{equation*}
for all \(x_1, x_2 \in X\text{.}\) We call \(L\) a Lipschitz constant for \(f\text{.}\)
It is clear that a uniformly continuous map is continuous, but a continuous map is not necessarily uniformly continuous. Moreover, Lipschitz continuity implies uniform continuity, but uniform continuity does not imply Lipschitz continuity.
Some of the other properties of continuous functions on an interval can be generalised to the setting of metric spaces, too.
Theorem 1.45. Continuity of compositions.
Suppose that \((X, d_X)\text{,}\) \((Y, d_Y)\text{,}\) and \((Z, d_Z)\) are metric spaces and that \(f \maps X \to Y\) and \(g \maps Y \to Z\) are maps between them. Let \(x_0 \in X\text{.}\) If \(f\) is continuous at \(x_0\) and \(g\) is continuous at \(f(x_0)\text{,}\) then \(g \circ
f\) is continuous at \(x_0\text{.}\)
Proof.
Fix \(\varepsilon \gt 0\text{.}\) As \(g\) is continuous at \(f(x_0)\text{,}\) there exists \(\rho \gt 0\) such that
\begin{equation*}
d_Y(y, f(x_0)) \lt \rho \implies d_Z(g(y), g(f(x_0))) \lt \varepsilon\text{.}
\end{equation*}
As \(f\) is continuous at \(x_0\text{,}\) there exists a \(\delta \gt 0\) such that
\begin{equation*}
d_X(x, x_0) \lt \delta \implies d_Y(f(x), f(x_0)) \lt \rho\text{.}
\end{equation*}
So if \(d_X(x, x_0) \lt \delta\text{,}\) it follows that \(d_Z(g(f(x)), g(f(x_0))) \lt \varepsilon\text{.}\)
Theorem 1.46. Sequential continuity.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Let \(x_0 \in X\) and let \(f \maps X \to Y\) be a map. The following are equivalent.
- The map \(f\) is continuous at \(x_0\text{.}\)
- If \((x_n)_{n \in \N}\) is a sequence in \(X\) with \(x_n \to x_0\) as \(n \to \infty\text{,}\) then \(f(x_n) \to f(x_0)\) as \(n \to \infty\text{.}\)
Proof.
Suppose that \(f\) is continuous at \(x_0\text{.}\) Let \((x_n)_{n \in
\N}\) be a sequence with \(x_n \to x_0\) as \(n \to \infty\text{.}\) Let \(\varepsilon \gt 0\) and choose \(\delta \gt 0\) such that
\begin{equation*}
d_X(x,x_0) \lt \delta \implies d_Y(f(x),f(x_0)) \lt \varepsilon\text{.}
\end{equation*}
There exists \(N \in \N\) such that \(d_X(x_n,x_0) \lt \delta\) whenever \(n \ge N\text{,}\) and so \(d_Y(f(x_n),f(x_0)) \lt \varepsilon\text{.}\) It follows that \(f(x_n) \to f(x_0)\) as \(n \to \infty\text{.}\)
Now suppose that \(f\) is not continuous at \(x_0\text{.}\) That is,
\begin{gather*}
\exists \varepsilon \gt 0
\st
\forall \delta \gt 0
\,
\exists x \in X
\st\\
d_X(x,x_0) \lt \delta
\anda
d_Y(f(x),f(x_0)) \ge \varepsilon\text{.}
\end{gather*}
Choose \(\varepsilon \gt 0\) with this property and consider \(\delta =
1/n\) for an arbitrary \(n \in \N\text{.}\) We conclude that there exists a point \(x_n \in X\) with \(d_X(x_n,x_0) \lt 1/n\) and \(d_Y(f(x_n),f(x_0)) \ge \varepsilon\text{.}\) Since we have such a point \(x_n\) for every \(n \in \N\text{,}\) we obtain a sequence \((x_n)_{n \in
\N}\text{.}\) This sequence satisfies \(x_n \to x_0\text{,}\) but \(f(x_n) \not\to
f(x_0)\) as \(n \to \infty\text{.}\)
The second condition in Theorem 1.46 is called sequential continuity. Often it is much more convenient to check sequential continuity than to directly verify Definition 1.42, which makes Theorem 1.46 a very useful result.
Proposition 1.47. Continuity of basic operations.
- If \((X,d)\) is a metric space, then \(d \maps X \times X \to \R\) is continuous.
- If \((X,\n\blank)\) is a normed space, then \(\n\blank \maps X \to \R\) is continuous, as are vector addition and scalar multiplication, thought of as maps \(a \maps X \times X \to X\) and \(s \maps \R \times X \to X\) defined by \(a(x,y)=x+y\) and \(s(\alpha,x) = \alpha x\text{.}\)
- If \((X,\scp\blank\blank)\) is an inner product space, then \(\scp\blank\blank \maps X \times X \to \R\) is continuous.
Proof.
By Theorem 1.46, it is enough to check for sequential continuity, which follows from Exercise 1.3.1 and Exercise 1.3.2.
Exercises Exercises
1. (PS4) Metric is Lipschitz.
Let \((X,d)\) be a metric space.
(a)
Fix \(p \in X\text{.}\) Show that the function \(f \maps X \to \R\text{,}\) given by \(f(x) = d(x, p)\) for \(x \in X\text{,}\) is Lipschitz continuous.
Hint.
The Reverse triangle inequality is useful here.
(b)
Consider the product space \(X \times X\) equipped with the product metric \(d_{X \times X}\text{.}\) Show that \(d \maps X \times X \to \R\) is Lipschitz continuous with respect to \(d_{X \times X}\text{.}\)
Hint 1.
A substantial portion of this problem is translating the statement “\(d
\maps X \times X \to \R\) is Lipschitz continuous” into an inequality to be proven. It is worth your time to do this translation carefully!
Hint 2.
The obvious inequality \(d(x_1,x_2) \le d_{X \times X}((x_1,y_1),(x_2,y_2))\) is useful here, as is the estimate used in the first part of Exercise 1.3.1.
2. (PS4) Image of closure under continuous map.
Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces, suppose that \(f
\maps X \to Y\) is a continuous map, and let \(S \subseteq X\text{.}\) Show that the images \(f(\overline S)\) and \(f(S)\) are related by \(f(\overline S) \subseteq \overline{f(S)}\text{.}\)
Hint.
Probably the simplest thing, given the tools available, is to argue directly using the definition of the closure in Definition 1.20 and the definition of continuity. Alternatively, one can argue in terms of convergent sequences, partly emulating the proof of Theorem 1.35.
3. Continuity in terms of open sets.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and \(f \maps X \to
Y\text{.}\) Show that \(f\) is continuous if and only if the preimage \(f^{-1}(U)\) is open for all open subsets \(U\) of \(Y\text{.}\)
