Section Introduction
Recall that we could characterize a function \(f\) from a metric space \((X,d_X)\) to a metric space \((Y,d_Y)\) as continuous at \(a \in X\) if \(f^{-1}(N)\) is a neighborhood of \(a\) in \(X\) whenever \(N\) is a neighborhood of \(f(a)\) in \(Y\text{.}\) We have defined neighborhoods in topological spaces, so we can use this characterization as our definition of a continuous function from one topological space to another.
Definition 14.1.
A function \(f\) from a topological space \((X, \tau_X)\) to a topological space \((Y, \tau_Y)\) is continuous at a point \(a \in X\) if \(f^{-1}(N)\) is a neighborhood of \(a\) in \(X\) whenever \(N\) is a neighborhood of \(f(a)\) in \(Y\text{.}\) The function \(f\) is continuous if \(f\) is continuous at each point in \(X\text{.}\)
We saw that in metric spaces, a useful characterization of continuity was in terms of open sets. It is not surprising that we have the same characterization in topological spaces. You may assume the result of
Theorem 14.2 (the topological space version of
Theorem 8.5 for metric spaces) for this activity.
Theorem 14.2.
Let \(f\) be a function from a topological space \((X,\tau_X)\) to a topological space \((Y,\tau_Y)\text{.}\) Then \(f\) is continuous if and only if \(f^{-1}(O)\) is an open set in \(X\) whenever \(O\) is an open set in \(Y\text{.}\)
Preview Activity 14.1.
(a)
Let
\begin{equation*}
(X, \tau_X) = (\{1,2,3,4\}, \{\emptyset, \{1\}, \{2\}, \{1,2\}, X \})
\end{equation*}
and let
\begin{equation*}
(Y, \tau_Y) = (\{2,4,6,8\}, \{\emptyset, \{4\}, \{6\}, \{4,6\}, Y\})\text{.}
\end{equation*}
Define \(f: X \to Y\) by \(f(x) = 2x\text{.}\)
(i)
Is \(f\) continuous at \(4\text{?}\)
(ii)
Is \(f\) a continuous function?
(b)
Let
\begin{equation*}
(X, \tau_X) = (\{1,2,3,4\}, \{\emptyset, \{1\}, \{2\}, \{1,2\}, \{3,4\}, X \})
\end{equation*}
and let
\begin{equation*}
(Y, \tau_Y) = (\{a,b,c\}, \{\emptyset, \{a\}, \{a,c\}, Y\})\text{.}
\end{equation*}
Define \(f: X \to Y\) by \(f(1) = a\text{,}\) \(f(2) = c\text{,}\) \(f(3) = f(4) = b\text{.}\)
(i)
Show that \(f\) is a continuous function.
(ii)
Even though \(f\) is continuous, it is possible that \(f(O)\) may not be open for every open set in \(X\text{.}\) Find such an example for this function \(f\text{.}\)
(c)
Functions \(f\) that have the property that \(f(O)\) is open whenever \(O\) is open in \(X\) are called open functions.
Definition 14.3.
Let \(f: X \to Y\) be a function from a topological space \(X\) to a topological space \(Y\text{.}\) Then \(f\) is an open function if \(f(U)\) is open in \(Y\) whenever \(U\) is open in \(X\text{.}\)
There is a similar definition of a closed function.
(d)
Let \(X= \{1,2,3,4,5\}\) and \(\tau = \{\emptyset,\{1\}, \{3,5\}, \{1,3,5\}, X\}\text{.}\) Define \(f: X \to X\) by \(f(x) = \la x-3 \ra+1\text{.}\) At which points is \(f\) continuous? Is \(f\) a continuous function?
(e)
Let \(f:(\Z,\tau_{FC}) \to (\Z, d_E)\) where \(f(n) = n\) and \(\tau_{FC}\) is the finite complement topology. Is \(f\) a continuous function? If \(f\) is not continuous, exhibit a specific point at which \(f\) fails to be continuous. Explain.
(f)
Let \(f:(\Z, d_E) \to (\Z,\tau_{FC})\) where \(f(n) = n\) and \(\tau_{FC}\) is the finite complement topology. Is \(f\) a continuous function? If \(f\) is not continuous, exhibit a specific point at which \(f\) fails to be continuous. Explain.
(g)
It can sometimes be easier to show that a function \(f\) mapping a topological space \((X,d_X)\) to a topological space \((Y,d_Y)\) is continuous by working with a basis instead of all open sets. Let \(\B\) be a basis for the topology on \(Y\text{.}\) Is it the case that if \(f^{-1}(B)\) is open for every \(B \in \B\text{,}\) then \(f\) is continuous? Verify your result.
To complete the introduction to this section, we prove
Theorem 14.2. We prove one direction now and leave the other for the next activity.
Let \(f\) be a function from a topological space \((X, \tau_X)\) to a topological space \((Y, \tau_Y)\text{.}\) We first assume that \(f\) is continuous and show that \(f^{-1}(O)\) is an open set in \(X\) whenever \(O\) is an open set in \(Y\text{.}\) Suppose that \(O\) is an open set in \(Y\text{.}\) To show that \(f^{-1}O)\) is open in \(X\text{,}\) we will show that \(f^{-1}(O)\) is a neighborhood of each of its points. Let \(a \in f^{-1}(O)\text{.}\) Then \(f(a) \in O\text{.}\) Since \(O\) is an open set, \(O\) is a neighborhood of \(f(a)\text{.}\) The fact that \(f\) is continuous means that \(f^{-1}(O)\) is a neighborhood of \(a\text{.}\) So \(f^{-1}(O)\) is a neighborhood of each of its points and \(f^{-1}(O)\) is an open set.
Activity 14.2.
Now we prove the remaining implication in
Theorem 14.2. That is, let
\(f\) be a function from a topological space
\((X, \tau_X)\) to a topological space
\((Y, \tau_Y)\text{,}\) and assume that
\(f^{-1}(O)\) is open whenever
\(O\) is open in
\(Y\text{.}\) We will prove that
\(f\) is a continuous function.
(a)
Using the definition, what does it take to show that \(f\) is a continuous function?
(b)
Let \(a \in X\) and suppose that \(N\) is a neighborhood of \(f(a)\) in \(Y\text{.}\) What can we conclude from \(N\) being a neighborhood?
(c)
Use the assumption about \(f\) in this activity to explain why \(f^{-1}(N)\) is a neighborhood of \(a\) in \(X\text{.}\)
(d)
Explain how we have shown that \(f\) is a continuous function.
Theorem 14.4.
Let \(f\) be a function from a topological space \((X,\tau_X)\) to a topologicval space \((Y,\tau_Y)\text{.}\) Then \(f\) is continuous if and only if \(f^{-1}(C)\) is closed in \(X\) whenever \(C\) is a closed set in \(Y\text{.}\)
