In calculus you learn that a function \(f\colon \R\rightarrow \R\) is defined to be continuous if for all \(\epsilon > 0\) there is a \(\delta > 0\) such that for all \(x,x'\in \R\) we have the implication
Now that we have some experience generalizing analysis concepts to topological ones, let’s see how best this translates to arbitrary topologies.
Definition1.9.1.Continuous function.
Let \((X,\mathcal{T}), (Y,\mathcal{T}')\) be topological spaces. A function \(f\colon X\rightarrow Y\) is continuous if for all open sets \(U\subseteq Y\text{,}\) the set \(f^{-1}(U)\) is open in \(X\text{.}\) Using logical shorthand:
Continuity is defined in terms of preimages of sets under a function \(f\text{.}\) Thankfully, the preimage operation behaves very well with respect to our various set operations. In particular, we have the following identities, which you may make use of without further ado:
Given a topological space \((X,\mathcal{T})\text{,}\) the identity function
\begin{align*}
\id_X\colon X \amp \rightarrow X\\
x\amp\mapsto x
\end{align*}
is continuous. Here we take \(\mathcal{T}\) as the topology of both the domain and codomain.
Constant functions.
Let \(X, Y\) be topological spaces. For any fixed \(y_0\in Y\) the constant function \(f\text{,}\) defined as \(f(x)=y_0\) for all \(x\in X\) is continuous.
Subspace inclusions.
Let \(Y\) be a subspace of \(X\text{.}\) The inclusion map \(i\colon Y\hookrightarrow X\) is continuous with respect to the subspace topology on \(Y\text{.}\)
Maps from discrete spaces.
If \(X\) is equipped with the discrete topology, then any function \(f\colon X\rightarrow Y\) to a topological space \(Y\) is continuous.
Maps to trivial spaces.
If \(Y\) is equipped with the trivial topology, then any map \(f\colon X\rightarrow Y\) from a topological space to \(Y\) is continous.
Solution.
These are all pretty elementary. I’ll prove (3). Let \(Y\subseteq X\) be a subspace and let \(i\colon Y\hookrightarrow X\) continuous. Given any open set \(U\subseteq X\text{,}\) we have \(i^{-1}(U)=U\cap Y\) which by definition is open in the subspace topology. Thus \(i\) is continuous.
Example1.9.4.Non-continuous inclusion.
Show that the identity map \(\id_\R\colon \R\hookrightarrow \R_l\) is not continuous. Here the topology of the domain is the standard one, and the topology on the codomain is the lower limit topology.
Solution.
For any \(U\subseteq \R\) we have \(\id_\R^{-1}(U)=U\text{.}\) If we choose \(U\) to be any open set of \(\R_l\) that is not open in the standard topology, it follows that \(\id_{\R}^{-1}(U)\) is not open in \(\R\) and hence that \(\id_\R\) is not continuous. We may take \(U=[0,1)\) for example.
Note: let \(\mathcal{T}_1, \mathcal{T}_2\) be two topologies on the set \(X\text{,}\) and for \(1\leq i\leq 2\) let \(X_i\) denote the space with underlying set \(X\) and topology \(\mathcal{T}_i\text{.}\) The same argument above can be used to show that \(\id_X\colon X_1\rightarrow X_2\) is continuous if and only if \(\mathcal{T}_2\subseteq \mathcal{T}_1\text{.}\)
Example1.9.5.Products and projections.
Let \(\{X_i, \mathcal{T}_i\}_{i\in I}\) be a collection of topological spaces, and let \(X=\prod_{i\in I}X_i\text{.}\) For each \(j\in I\) define the projection function \(\pi_j\colon X\rightarrow X_j\) as \(\pi_j((x_i)_{i\in I})=x_i\text{.}\) In other words, the \(j\)-th projection map returns the \(j\)-th coordinate of an element of \(X\text{.}\)
Prove: if \(X\) is given the product topology, then \(\pi_j\) is continuous for all \(j\in J\text{.}\)
Prove: given a topology \(\mathcal{T}\) on \(X\text{,}\) we have \(\pi_j\) continuous for all \(j\in J\) if and only if \(\mathcal{T}\) contains the product topology. Thus, in particular, the product topology is the coarsest topology on \(X\) making each projection map continuous.
Solution.
It is enough to prove (2), since it clearly implies (1). The maps \(\pi_j\) are continuous in a topology \(\mathcal{T}\) if and only if for all \(j\in I\) and open \(U_j\subseteq X_j\) the preimage \(\pi_j^{-1}(U_j)=U_j\times \prod_{i\ne j}X_i\) is open in \(\mathcal{T}\text{.}\) Next it clear that \(U_j\times \prod_{i\ne j}X_i\) is open in \(\mathcal{T}\) for all \(j\in J\) if and only if
is open for any finite set \(\{j_1, j_2,\dots, j_n\}\subseteq I\text{.}\) Since these sets form a basis for the product topology, we conclude that the projections are all continuous if and only if \(\mathcal{T}\) is finer than the product topology.
Theorem1.9.6.
Let \(f\colon X\rightarrow Y\) be a function between the topological spaces \(X, Y\text{,}\) and let \(\mathcal{B}\) be a basis for \(Y\text{.}\) The following statements are equivalent.
\(f\) is continuous.
For all \(B\in \mathcal{B}\text{,}\) the preimage \(f^{-1}(B)\) is open in \(X\text{.}\)
Proof.
It is clear that (1) implies (2). Assume (2) is true. Given open \(U\subseteq Y\text{,}\) we can write \(U=\bigcup_{i\in I}B_i\) with \(B_i\in \mathcal{B}\text{.}\) It follows that
is open, since by assumption each \(f^{-1}(B_i)\) is open.
Theorem1.9.7.Continuity equivalences.
Let \(f\colon X\rightarrow Y\) be a function between the topological spaces \(X, Y\text{.}\) The following statements are equivalent.
\(f\) is continuous.
For all \(A\subseteq X\text{,}\) we have \(f(\overline{A})\subseteq \overline{f(A)}\text{.}\)
For all closed sets \(C\subseteq Y\text{,}\) the preimage \(f^{-1}(C)\) is closed in \(X\text{.}\)
Continuity at \(x\).
For all \(x\in X\) and all sets \(V\subseteq Y\) containing \(y=f(x)\text{,}\) there is an open set \(U\subseteq X\) containing \(x\) such that \(f(U)\subseteq V\text{.}\)
Proof.
\((1)\implies (2)\text{.}\) Take \(y\in f(\overline{A})\text{.}\) We have \(y=f(x)\) for some \(x\in \overline{A}\text{.}\) If \(V\) is any open neighborhood of \(y\text{,}\) then since \(f\) is continuous \(U=f^{-1}(V)\) is an open neighborhood of \(x\text{.}\) Since \(x\in\overline{A}\text{,}\) there exists \(x'\in A\cap U\text{.}\) It follows that \(y'=f(x')\in V\cap f(A)\text{,}\) showing this intersection is nonempt. We concdlue that \(y\in \overline{f(A)}\text{,}\) as desited.
\((2)\implies (3)\text{.}\) Assume (2) holds of \(f\text{.}\) Given closed \(C\subseteq Y\) let \(A=f^{-1}(C)\text{.}\) To show \(A\) is closed, we will show that \(\overline{A}\subseteq A\text{.}\) Since \(A=f^{-1}(C)\text{,}\) this is equivalent to showing \(f(\overline{A})\subseteq C\text{;}\) but this follows from (2) since \(f(\overline{A})\subseteq \overline{f(A)}\subseteq C\text{,}\) where the last inclusion in this chain follows from the fact that \(C\) is closed and contains \(f(A)\text{.}\)
\((3)\implies (4)\text{.}\) Let \(x\in X\text{,}\) and let \(V\) be an open neighborhood of \(y=f(x)\text{.}\) The set \(C=Y-V\) is closed, whence \(f^{-1}(C)=X-f^{-1}(V)\) is closed. It follows that \(U=f^{-1}(V)\) is an open set containing \(x\text{,}\) and we have \(f(U)=f(f^{-1}(V))\subseteq V\text{.}\)
\((4)\implies (1)\text{.}\) We assume (4). Let \(V\subseteq Y\) be an open set. For any \(x\in f^{-1}(V)\text{,}\) there is by (4) (taking open set \(V\ni y=f(x)\)) an open set \(U_x\ni x\) such that \(f(U_x)\subseteq V\text{.}\) It follows that \(U_x\subseteq f^{-1}(V)\text{.}\) Thus for any \(x\in f^{-1}(V)\) there is an open \(U_x\ni x\) such that \(U_x\subseteq V\text{.}\) It follows that \(f^{-1}(V)\) is open, as desired.
Definition1.9.8.Continuity at \(x\).
Let \(f\colon X\rightarrow Y\) be a map between topological spaces. We say that \(f\) is continuous at x, for \(x\in X\text{,}\) if for all open neighborhoods \(V\ni y\text{,}\) there is an open neighborhood \(U\ni x\) such that \(f(U)\subseteq V\text{.}\)
Theorem1.9.9.Building continuous functions.
Composition.
If \(f\colon X\rightarrow Y\) and \(g\colon Y\rightarrow Z\) are continuous, then the composition \(g\circ f\colon X\rightarrow Z\) is continuous.
Restriction.
If \(f\colon X\rightarrow Y\) is continuous, and \(Z\subseteq X\) is a subspace, then the restriction \(f\vert_Z\colon Z\rightarrow Y\) is continuous.
Expansion.
If \(f\colon X\rightarrow Y\) is continuous and \(Y\) is a subspace of \(Z\text{,}\) then the function \(f\colon X\rightarrow Z\) obtained by composing \(f\) with the inclusion \(Y\hookrightarrow Z\) is continuous.
Local on source (open pasting).
If \(f\colon X \rightarrow Y\) is a function between topological spaces and there is a collection of open sets \(\{U_i\}_{i\in I}\) such that \(X=\bigcup_{i\in I}U_i\) and the restrictions \(f_i\colon U_i\rightarrow Y\) are continuous, then \(f\) is continuous.
Closed pasting.
Let \(X, Y\) be topological spaces, and suppose \(X=C\cup D\text{,}\) where \(C, D\) are closed sets. Given continuous functions \(f\colon C\rightarrow Y\) and \(g\colon D\rightarrow Y\) that satisfy \(f(x)=g(x)\) for all \(x\in C\cap D\text{,}\) the function \(h\colon X\rightarrow Y\) defined as
Statements (1)-(3) are elementary, and can be found in Munkres.
Statement (4) follows from equivalent condition (4) from Theorem 1.9.7. Indeed, take any \(x\in X\) and open set \(V\ni y=f(x)\text{.}\) Since \(x\in U_i\) for some \(i\in I\text{,}\) and since \(f_i\colon U_i\rightarrow Y\) is continuous, there is an open \(U\ni x\) with \(U\subseteq U_i\text{,}\) such that \(f(U)\subseteq V\text{.}\) Since \(U\) is open in \(U_i\) and \(U_i\) is open in \(X\text{,}\) we conclude \(U\) is open in \(X\text{,}\) as desired.
To prove statement (5) we show that \(h^{-1}(A)\) is closed for any closed set \(A\subseteq Y\text{.}\) Since \(h(x)\in A\) if and only if \(x\in C\) and \(f(x)\in A\) or \(x\in D\) and \(g(x)\in A\text{,}\) we have \(h^{-1}(A)=f^{-1}(A)\cup g^{-1}(A)\text{.}\) Since \(f\) and \(g\) are closed we have \(f^{-1}(A)\) closed in \(C\) and \(g^{-1}(A)\) closed in \(D\text{.}\) Since \(C\) and \(D\) are closed, transitivity of closedness implies \(f^{-1}(A)\) and \(g^{-1}(A)\) are closed in \(X\text{.}\) We conclude \(h^{-1}(A)=f^{-1}(A)\cup g^{-1}(A)\) is closed, as desired.
