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 \(gof:X\rightarrow Z\) 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 \(f^{-1}(g^{-1}(V))=(gof)^{-1}(V)\). Hence the composition gof is continuous.
Continuity of functions on topological spaces may also be stated in terms of closed sets.
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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 \(X-f^{1-}(G)=f^{-1}(Y-G)\). The result therefore follows immediately from the definitions
of continuity and closed sets.
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Theorem 2.3.4. Let \(X\) and \(Y\) be topological spaces and \(f:X\rightarrow Y\) be a function. Suppose that \(X=A_1\cup A_2\cup ........\cup A_n\), where \(A_1,A_2,.......,A_n\)
are closed sets in \(X\). Also suppose that the restriction of \(f\) to the sets \(A_i\) is continuous for \(i=1,2,.......,n\). 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 the function set
V under the restriction \(f/A_i\) of \(f\) to \(A_i\) is \(f^{-1}(V)\cap A_i\), \(\forall i=1,2,...........,n\). It follows that the continuity of \(f/A_i\) that \(f^{-1}(V)\cap A_i\) is open in \(A_i\), \(\forall i=1,2,......,n\). So there exists
open sets \(U_1,U_2,.............,U_n\) in \(X\) such that \(f^{-1}(V)\cap A_i=U_i\cap A_i\), \(\forall i=1,2,........,n\).
Let \(W_i=U_i\cup (X-A_i)\) for every \(i=1,2,.........,n\). 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 \(i=1,2,.........,n\).
We claim that \(f^{-1}(V)=W_1\cap W_2\cap .........\cap W_n\). Let \(W=W_1\cap W_2\cap .....\cap W_n\). Then \(f^{-1}(V)\subseteq W\), since \(f^{-1}(V)\subseteq W_i\) for each \(i=1,2,..........,n\). 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 \(X=A_1\cup A_2\cup ..................\cup A_n\) and \(W_i\cap A_i=f^{-1}(V)\cap A_i\) for each \(i=1,2,..........,n\).
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 \(A\).
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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 defined on
the interval [0,1] and [0,1], where
\(\alpha (1)=\beta (0)\).
\(\gamma :[0,1]\rightarrow Y\) be defined by \[ \gamma (t)= \begin {cases} \alpha (2t) & \text {if}\hspace {0.2cm} 0\leq t\leq \dfrac {1}{2}\\\\ \beta (2t-1) & \text {if}\hspace {0.2cm} \dfrac {1}{2}\leq t\leq 1 \end {cases} \]
Now \(\gamma \Big |_{\large [0,\frac {1}{2}]=\large \alpha o \rho }\) 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 \(\gamma \Big |_{\large [0,\frac {1}{2}]}\) is continuous been the
composition of two continuous functions.
Similarly \(\gamma \Big |_{\large [\frac {1}{2},1]}\) 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 \(\gamma :[0,1]\) 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 \(Id_X(V)=V\). This means that \(Id_X^{-1}(V)\) 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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