2.3 Continuous Functions

In this section we will study continuous functions on topological spaces. As we will see, continuity of a function on topological space is a generalization of continuity of a function in a metric space.

Definition 2.3.1. Let \(X\) and \(Y\) be topological spaces. A mapping \(f:X\rightarrow Y\) is said to be continuous if \(f^{-1}(V)\) is an open set in \(X\) for every open set \(V\) in \(Y\).

Theorem 2.3.2. Let \(X\,,\,Y\) and \(Z\) be topological spaces and let \(f:X\rightarrow Y\) and \(g:Y\rightarrow Z\) be continuous functions. Then the composition \(g\circ f:X\rightarrow Z\) of the functions \(f\) and \(g\) is continuous.

Proof. Let \(V\) be an open set in \(Z\). Then \(g^{-1}(V)\) is an open set in \(Y\) (since \(g\) is continuous) and hence \(f^{-1}(g^{-1}(V))\) is open in \(X\) (since \(f\) is continuous). But \(f^{-1}(g^{-1}(V))=(g\circ f)^{-1}(V)\). Hence the composition \(g\circ f\) is continuous. □

Continuity of functions on topological spaces may also be stated in terms of closed sets.

Theorem 2.3.3. Let \(X\) and \(Y\) be topological spaces. Then a function \(f:X\rightarrow Y\) is continuous if and only if \(f^{-1}(G)\) is closed in \(X\) for every closed subset \(G\) of \(Y\).

Proof. If \(G\) is any subset of \(Y\), then \(X\setminus f^{-1}(G)=f^{-1}(Y\setminus G)\). The result therefore follows immediately from the definitions of continuity and closed sets. □

Theorem 2.3.4. Let \(X\) and \(Y\) be topological spaces and \(f:X\rightarrow Y\) be a function. Suppose that \(X=A_1\cup A_2\cup \cdots \cup A_n\), where \(A_1,A_2,\ldots ,A_n\) are closed sets in \(X\). Also suppose that the restriction of \(f\) to the sets \(A_i\) is continuous for \(i=1,2,\ldots ,n\). Then \(f\) is continuous on \(X\).

Proof. Let \(V\) be any open set in \(Y\). We must show that \(f^{-1}(V)\) is open in \(X\). Now the pre-image of \(V\) under the restriction \(f|_{A_i}\) of \(f\) to \(A_i\) is \(f^{-1}(V)\cap A_i\), for \(i=1,2,\ldots ,n\). It follows from the continuity of \(f|_{A_i}\) that \(f^{-1}(V)\cap A_i\) is open in \(A_i\), for \(i=1,2,\ldots ,n\). So there exists open sets \(U_1,U_2,\ldots ,U_n\) in \(X\) such that \(f^{-1}(V)\cap A_i=U_i\cap A_i\) for \(i=1,2,\ldots ,n\). Let \(W_i=U_i\cup (X-A_i)\) for every \(i=1,2,\ldots ,n\). Then \(W_i\) is an open set in \(X\) (as it is a union of open sets \(U_i\) and \(X-A_i\)) and \(W_i\cap A_i=U_i\cap A_i=f^{-1}(V)\cap A_i\) for each \(i=1,2,\ldots ,n\). We claim that \(f^{-1}(V)=W_1\cap W_2\cap \cdots \cap W_n\). Let \(W=W_1\cap W_2\cap \cdots \cap W_n\). Then \(f^{-1}(V)\subseteq W\), since \(f^{-1}(V)\subseteq W_i\) for each \(i=1,2,\ldots ,n\). Also \begin {align*} W &=\bigcup ^n_{i=1}(W\cap A_i)\subseteq \bigcup ^n_{i=1}(W_i\cap A_i)\\ &=\bigcup ^n_{i=1}(f^{-1}(V)\cap A_i)\subseteq f^{-1}(V) \end {align*}

since \(X=A_1\cup A_2\cup \cdots \cup A_n\) and \(W_i\cap A_i=f^{-1}(V)\cap A_i\) for each \(i=1,2,\ldots ,n\). Therefore, \(f^{-1}(V)=W\). But \(W\) is open in \(X\), since it is the intersection of a finite collection of open sets. We have thus shown that \(f^{-1}(V)\) is open in \(X\), for any open set \(V\) in \(Y\). Thus \(f\) is continuous. □

In the following example we will need to recall that if \(f:X\rightarrow Y\) is a function and \(A\subseteq X\), then \(f|_A\) denotes the restriction of \(f\) to the set \(A\).

Example 2.3.5. Let \(Y\) be a topological space and let \(\alpha :[0,1]\rightarrow Y\) and \(\beta :[0,1]\rightarrow Y\) be continuous functions with \(\alpha (1)=\beta (0)\). Let \(\gamma :[0,1]\rightarrow Y\) be defined by \[ \gamma (t)= \begin {cases} \alpha (2t) & \text {if}\hspace {0.2cm} 0\leq t\leq \frac {1}{2}\\ \beta (2t-1) & \text {if}\hspace {0.2cm} \frac {1}{2}\leq t\leq 1 \end {cases} \]

Now \(\gamma \big |_{[0,\frac {1}{2}]}=\alpha \circ \rho \) where \(\rho :[0,\frac {1}{2}]\rightarrow [0,1]\) is the continuous function defined by \(\rho (t)=2t\) for all \(t\in [0,\frac {1}{2}]\). Thus \(\gamma \big |_{[0,\frac {1}{2}]}\) is continuous, being the composition of two continuous functions. Similarly \(\gamma \big |_{[\frac {1}{2},1]}\) is continuous. The sub-intervals \([0,\frac {1}{2}]\) and \([\frac {1}{2},1]\) are closed in [0,1], and [0,1] is the union of these two sub-intervals. It follows from the previous theorem that \(\gamma :[0,1]\rightarrow Y\) is continuous.

Proposition 2.3.6. Let \(X\) and \(Y\) be topological spaces. Then

i: the identity map \(Id_X:X\rightarrow X\) is continuous.
ii: any constant map \(X\) to \(Y\) is continuous.

Proof.

i: Let \(V\) be any open set in \(X\). Then \(Id_X^{-1}(V)=V\), which is open in \(X\), so that \(Id_X\) is continuous.
ii: Let \(K:X\rightarrow Y\) be the constant map defined by \(K(x)=y_0\) for all \(x\in X\) and for some \(y_0\in Y\). Let \(V\) be an open set in \(Y\). If \(y_0\not \in V\), then \(K^{-1}(V)=\emptyset \). If \(y_0\in V\), then \(K^{-1}(V)=X\). Since \(\emptyset \) and \(X\) are both open sets in \(X\,,\, K\) is continuous.

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