1.6 Continuous Functions On Metric Spaces
In previous courses you have studied continuity of real valued functions.
In this section we will study continuity of functions between general metric spaces.
Definition 1.6.1. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A function \(f\) is said to be continuous at a point \(x_0\)
in \(X\) if given \(\epsilon >0\), there exists a \(\delta >0\) such that \(d_Y(f(x),f(x_0))<\epsilon \) whenever \(d_X(x,x_0)<\delta \).
We say that \(f\) is continuous on a subset \(A\) of \(X\) if it is continuous at every point of \(A\).
Example 1.6.2. Consider the metric space \((\mathbb {R},d)\), where \(d\) is the usual metric on \(\mathbb {R}\) and \(f:\mathbb {R} \rightarrow \mathbb {R}\) is the function
defined by \(f(x)=x^2\). Then \(f\) is continuous at the point \(x=0\).
Proof. Let \(\epsilon >0\) be given and let x be an arbitrary point of \(\mathbb {R}\). Then we have \(d(f(x),f(x_0))=|f(x)-f(x_0)|=|x^2|=|x|^2\). If we take \(\delta =\sqrt {\epsilon }\), we obtain \(|x-0|=|x|<\delta \)
implies \(|f(x)-f(x_0)|=|x|^2<\delta ^2 =\epsilon \).
The next theorem characterises continuity of a function at a point in a metric space in terms
of open balls.
□
Theorem 1.6.3. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Then a function \(f:X\rightarrow Y\) is continuous at a point \(x_0\) in \(X\) if
and only if given any open ball \(B(f(x_0),\epsilon )\) in \(Y\), there exists an open ball \(B(x_0,\delta )\) in \(X\) such that
\(f(B(x_0,\delta ))\subseteq B(f(x_0),\epsilon )\).
Proof. Suppose that \(f\) is continuous at \(x_0\) in X. Then for every \(\epsilon >0\) there exists a \(\delta >0\) such that \(d_Y(f(x),f(x_0))<\epsilon \) whenever
\(d_X(x,x_0)<\delta \). In other words, \(f(x)\in B(f(x_0),\epsilon )\) whenever \(x\in B(x_0,\delta )\). This implies that \(f(B(x_0,\delta ))\subseteq B(f(x_0),\epsilon )\).
Conversely, suppose that for every \(\epsilon >0\) there exists a \(\delta >0\) such that \(f(B(x_0,\delta ))\subseteq B(f(x_0),\epsilon )\). This means that \(f(x)\in B(f(x_0),\epsilon )\) whenever \(x\in B(x_0,\delta )\). In
other words, \(d_Y(f(x),f(x_0))<\epsilon \) whenever \(d_X(x,x_0)<\delta \). Hence f is continuous at \(x_0\).
The next theorem provides a criterion for continuity of functions between metric spaces in terms
of open sets.
□
Theorem 1.6.4. Let \(f:X\rightarrow Y\) be a function between metric spaces. Then \(f\) is continuous if and only if \(f^{-1}(U)\)
is open in \(X\) whenever \(U\) is open in \(Y\).
Proof. Suppose that \(f\) is continuous and \(U\) is open in \(Y\). Let \(x\) be an arbitrary element of \(f^{-1}(U)\). Then \(f(x)\in U\)
and since \(U\) is open, there is an \(\epsilon >0\) such that \(B(f(x),\epsilon )\subseteq U\). If \(f\) is continuous, the previous theorem implies that
there exists a \(\delta >0\) such that \(f(B(x,\delta ))\subseteq B(f(x),\epsilon )\) from \(f^{-1}(B(f(x),\epsilon ))\subseteq f^{-1}(U)\), we obtain that \(f^{-1}(U)\) is open in \(X\).
Conversely, suppose that \(f^{-1}(U)\) is open in \(X\) whenever \(U\) is open \(Y\). Let \(x\) be an arbitrary element of \(X\) and
let \(B(f(x),\epsilon )\subseteq Y\). Since \(B(f(x),\epsilon )\) is an open set in \(Y\), we have \(f^{-1}(B(f(x),\epsilon ))\) is open in \(X\). Now since
\(x\in f^{-1}(B(f(x),\epsilon ))\) there is \(\delta >0\) such that \(B(x,\delta )\subseteq f^{-1}(B(f(x),\epsilon ))\). Thus \(f(B(x,\delta ))\subseteq B(f(x),\epsilon )\). Hence f is continuous by the previous theorem.
Continuity of functions on metric spaces can also be characterised in terms of closed sets as the
next theorem shows.
□
Theorem 1.6.5. A function \(f:X\rightarrow Y\) between metric spaces is continuous if and only if \(f^{-1} (F)\) is closed in \(X\)
whenever \(F\) is closed in \(Y\).
Proof. Since \(F\) is closed in \(Y\), \(Y-F\) is open in \(Y\). Continuity of \(f\) then implies that \(f^{-1}(Y-F)\) is open in \(X\). But \(f^{-1}(Y-F)=f^{-1}(Y)-f^{-1}(F)=X-f^{-1}(F)\) and
so \(f^{-1}\) is closed in \(X\).
Conversely, suppose that \(f^{-1}(F)\) is closed in \(X\) whenever \(F\) is closed in \(Y\). Let \(U\) be an open set in \(Y\). Then \(Y-U\) is
closed in \(Y\) and so
\(f^{-1}(Y-U)=f^{-1}(Y)-f^{-1}(U)=X-f^{-1}(U)\) is closed in \(X\). Therefore \(f^{-1}(U)\) is open in \(X\). Since \(U\) was an arbitrary open set in Y, we obtain that \(f\) is
continuous.
□
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.