General Topology/Continuity

Definition (continuity):

Let $X,Y$ be topological spaces and let $f:X\to Y$ be a function. $f$ is called continuous if and only if for every open $U\subseteq Y$, the set $f^{-1}(U)$ is open.

Proposition (characterisation of continuity via subbasis):

Let $f:X\to Y$ be a function between topological spaces $X,Y$, and let ${\mathcal {B}}$ be a subbasis of the topology of $Y$. If $f^{-1}(V)$ is open in $X$ for all $V\in {\mathcal {B}}$, then $f$ is continuous.

Proof: Let $W\subseteq Y$ be an arbitrary open set; we are to show that $f^{-1}(W)$ is open. Now a basis of the topology on $Y$ is given by

${\mathcal {B}}':=\{B_{1}\cap \cdots \cap B_{n}|n\in \mathbb {N} ,B_{1},\ldots ,B_{n}\in {\mathcal {B}}\}$.

Since intersections commute with preimages, $f^{-1}(V)$ is open for $V\in {\mathcal {B}}'$. But all open sets of $Y$ are unions of elements of ${\mathcal {B}}'$, and preimages commute with unions, so that $f$ is continuous. $\Box$

Example (identity is a continuous function):

Let $X$ be a set with a topology $\tau$, and let $1_{X}$ denote the identity function on $X$. Then $1_{X}$ is continuous.

Proposition (category of topological spaces):

The class of all topological spaces, together with the continuous functions between them as morphisms, forms a category.

Proof: Indeed, the composition of continuous functions is again continuous, and further, the identity (which is unique, by composing any other identity with the above identity) is well-defined. $\Box$

Proposition (characterisation of continuity of functions to a space with initial topology):

Let $X$ be a set equipped with the initial topology of certain functions $f_{\alpha }:X\to Y_{\alpha }$, where $(Y_{\alpha })_{\alpha \in A}$ is a family of topological spaces. Let $Z$ be another topological space. A function $g:Z\to X$ is continuous if and only if all the functions $f_{\alpha }\circ g$ ($\alpha \in A$) are continuous.

Proof: First suppose that all the functions $f_{\alpha }\circ g$ are continuous, and let $U\subseteq X$ be open, so that we may write

$U=\bigcup _{\beta \in B}C_{\beta }$, $C_{\beta }=f_{\alpha _{\beta ,1}}^{-1}(U_{\alpha _{\beta ,1}})\cap \cdots \cap f_{\alpha _{\beta ,n_{\beta }}}^{-1}(U_{\alpha _{\beta ,n_{\beta }}})$

since we saw that the sets $C_{\beta }$ thus defined form a basis of the initial topology. Then $g^{-1}(U)$ is open in $Z$ since taking preimages commutes with taking unions and intersections. Further, observe that all functions $f_{\alpha }$ are continuous, so that the function $f_{\alpha }\circ g$ are continuous when $g$ is. $\Box$

Proposition (product of topological spaces is categorical product):

Let $(X_{\alpha })_{\alpha \in A}$ be topological spaces, and let $X$ be their product with product topology. Then $X$, together with the canonical projections $\pi _{\alpha }:X\to X_{\alpha }$, is a product of $(X_{\alpha })_{\alpha \in A}$ in the category of topological spaces and continuous functions.

Proof: Let another cone over the diagram be given that contains as points the $X_{\alpha }$ and no morphism, and suppose that the object of that cone is called $a$ and the arrows of that cone are called $\phi _{\alpha }:a\to X_{\alpha }$, so that $a$ is a topological space and the $\phi _{\alpha }$ are continuous functions. To ease notation, set $P:=\prod _{\alpha \in A}X_{\alpha }$, and define

$\psi :a\to P,\psi (x):=(\phi _{\alpha }(x))_{\alpha \in A}$.

Note that $\psi$ is the unique map that has the property that $\pi _{\alpha }\circ \psi (x)=\phi _{\alpha }(x)$, since the projection is simply given by taking the $\alpha$-th component. Furthermore, note that $\psi$ is continuous, since the $\phi _{\alpha }=\pi _{\alpha }\circ \psi$ are continuous, $P$ has the initial topology of the $\pi _{\alpha }$, and we conclude by the characterisation of continuity of functions to a space with initial topology. Thus, $P$ is a product. $\Box$

Continuity is a local property, in that it may be characterized by a property that a function might have at every point.

Definition (continuity at a point):

Let $X,Y$ be topological spaces, $x_{0}\in X$ a point, and $f:X\to Y$ a function. $f$ is called continuous at $x_{0}$ iff for every open neighbourhood $V\subseteq Y$ of $f(x_{0})$, there exists an open neighbourhood $U$ of $x_{0}$ such that $f(U)\subseteq V$.

Proposition (continuity is equivalent to continuity at each point):

Let $X,Y$ be topological spaces and $f:X\to Y$ be a function. $f$ is continuous if and only if it is continuous at all $x\in X$.

Proof: Suppose first that $f$ is continuous, and let $x\in X$. Let $V\subseteq Y$ be an open neighbourhood of $f(x)$, then by continuity $f^{-1}(V)=:U$ is an open neighbourhood of $x$ and by definition of the preimage $f(U)\subseteq V$. Suppose now that $f$ is continuous at each point $x\in X$, and let $V\subseteq Y$ be any open set. We are to show that $f^{-1}(V)$ is open. Indeed, let $x\in f^{-1}(V)$ be arbitrary, so that $f(x)\in V$. By continuity at $x$, we find that the sets $S_{x}$ of neighbourhoods $U$ of $x$ such that $f(U)\subseteq V$ is nonempty. (Note: We are inserting here an additional step of choosing a canonical neighbourhood for each $x$ in order to avoid the axiom of choice.) Define

$U_{x}:=\bigcup _{U\in S_{x}}U$,

which is an open neighbourhood of $x$ and has the property that $f(U_{x})\subseteq V$, that is, $U_{x}\subseteq f^{-1}(V)$. Then we obtain

$f^{-1}(V)\subseteq \bigcup _{x\in f^{-1}(V)}U_{x}\subseteq f^{-1}(V)$, ie. $f^{-1}(V)=\bigcup _{x\in f^{-1}(V)}U_{x}$,

which is open as the union of open sets. $\Box$

Proposition (characterisation of continuity by closed sets):

Let $f:X\to Y$ be a function between topological spaces.

$f$ is continuous if and only if $f^{-1}(A)$ is closed in $X$ for all closed subsets $A\subseteq Y$.

Proof: $f$ is continuous if and only if for all open subsets $V\subseteq Y$, the set $f^{-1}(V)$ is open in $X$. The open subsets of $Y$ are in bijective correspondence to the closed subsets of $Y$ via the map $V\mapsto Y\setminus V$, and the same holds for $X$. Now note that $f^{-1}(Y\setminus V)=f^{-1}(Y)\setminus f^{-1}(V)=X\setminus f^{-1}(V)$, so that the latter is closed precisely when the former is. In particular, the latter is always closed whenever $f^{-1}(V)$ is always open, and if the latter is always closed, then $f^{-1}(V)$ is always open. $\Box$

Proposition (continuity on closed union):

Let $X,Y$ be topological spaces, let $A,B\subseteq X$ be closed subsets of $X$ such that $A\cup B=X$ and let $f:X\to Y$ be a function such that $f|_{A}$ and $f|_{B}$ are both continuous. Then $f$ is continuous.

Proof: We show that $f^{-1}(C)$ is closed for all closed subsets $C\subseteq Y$, thereby proving continuity by the characterisation of continuous functions by the preimages of closed sets. Indeed, let $C\subseteq Y$ be closed. Note that

$f^{-1}(C)=f|_{A}^{-1}(C)\cup f|_{B}^{-1}(C)$,

that is, $f^{-1}(C)$ is the union of two closed sets and thus itself closed. $\Box$

Proposition (restriction of continuous function is continuous):

Let $X,Y$ be topological spaces, $S\subseteq X$ a subset and $f:X\to Y$ a continuous function. Then $f|_{S}$ is a continuous function from $S$ (with the subspace topology) to $Y$.

Proof: Let $V\subseteq Y$ be open. Then $f^{-1}(V)$ is open in $X$, so that $f^{-1}(V)\cap S=f|_{S}^{-1}(V)$ is open in the subspace topology on $S$. $\Box$

Definition (homeomorphism):

Let $X,Y$ be topological spaces, and let $f:X\to Y$ be a function which is invertible, ie. bijective. $f$ is called a homeomorphism if and only if both $f$ and $f^{-1}$ are continuous.

Proposition (restriction of homeomorphism is homeomorphism):

Let $X,Y$ be topological spaces and $f:X\to Y$ a homeomorphism. Let further $S\subseteq X$ be a subset. Then $f|_{S}:S\to f(S)$ is a homeomorphism, where $S$ and $f(S)$ bear the subspace topologies induced by $X$ resp. $Y$.

Proof: $f|_{S}$ is continuous since the restriction of a continuous function is continuous. Further, $f|_{S}^{-1}=f^{-1}|_{f(S)}$, which is likewise continuous as the restriction of a continuous map. Therefore, $f|_{S}$ is a homeomorphism. $\Box$

Definition (equicontinuity):

Let $X,Y$ be topological spaces. A subset $H\subseteq Y^{X}$ (where $Y^{X}$ denotes the set of all functions from $X$ to $Y$) is called equicontinuous at a point $x_{0}\in X$ iff for each $y\in Y$ and each open neighbourhood $V$ of $y$, there exist open neighbourhoods $U$ of $x_{0}$ and $W$ of $y$ so that for all $f\in H$ whenever $f(U)\cap W\neq \emptyset$, then $f(U)\subseteq V$. $H$ is called equicontinuous iff it is equicontinuous at each point.

Proposition (function in equicontinuous set is continuous):

Let $X,Y$ be topological spaces, let $H\subseteq Y^{X}$ be equicontinuous at $x_{0}\in X$ and let $f\in H$. Then $f$ is continuous at $x_{0}$.

Proof: Set $y:=f(x_{0})$. By definition of equicontinuity, whenever $V$ is a neighbourhood of $y$, we find neighbourhoods $U$ of $x_{0}$ and $W$ of $y$ so that whenever $g(U)\cap W\neq \emptyset$ for $g\in H$ arbitrary, then $g(U)\subseteq V$. But since $f(x_{0})=y$, we have $f(U)\cap W\neq \emptyset$, so that $f(U)\subseteq V$ and $f$ is continuous at $x_{0}$. $\Box$

When $Y$ is a uniform space, the definition of equicontinuity simplifies, and furthermore in this situation equicontinuous subsets are related to compact subsets of ${\mathcal {C}}(X,Y)$. This we will see in the chapter on uniform structures.

Definition (local homeomorphism):

Let $X,Y$ be topological spaces. A local homeomorphism is a function $f:X\to Y$ such that for all $x_{0}\in X$, we find an open neighbourhood $U$ of $x_{0}$ so that $f|_{U}$ is a homeomorphism and $f(U)$ is an open subset of $Y$.

In other words, a function $f:X\to Y$ is a local homeomorphism if and only if for all $x_{0}\in X$, there exists an open neighbourhood $U$ of $x_{0}$ and an open neighbourhood $V$ of $f(x_{0})$ so that $f|_{U}$ is a homeomorphism from $U$ to $V$.

Proposition (local homeomorphisms are open):

Let $f:X\to Y$ be a local homeomorphism. Then $f$ is an open map.

Proof: Indeed, let $U\subseteq X$ be open, and let $y\in f(X)$ be arbitrary. Pick $x\in f^{-1}(y)$ arbitrary. Then since $f$ is a local homeomorphism, we find $U\subseteq X$ open with $x\in U$ such that $f(U)$ is open. Further, $y\in f(U)\subseteq f(X)$. Since $\in f(X)$ was arbitrary, $f(X)$ is open. $\Box$

Definition (embedding):

An embedding is a continuous function $f:X\to Y$ between topological spaces so that $f$ is a homeomorphism between $X$ and $f(X)$.

Proposition (continuity criterion for the greatest lower bound topology):

Let $X$ be a set, let $(\tau _{\alpha })_{\alpha \in A}$ be a family of topologies on $x$, and suppose that $\tau$ is the greatest lower bound topology of $(\tau _{\alpha })_{\alpha \in A}$ on $X$. Let $f:X\to Y$ be a function, where $Y$ is a topological space. Then $f$ is continuous if and only if $f$ is continuous with respect to all topologies $\tau _{\alpha }$ ($\alpha \in A$) on $X$.

Proof: $f:X\to Y$ is continuous iff for all open $V\subseteq Y$, the set $f^{-1}(V)$ is open in $X$. This in turn is equivalent to $f^{-1}(V)$ being open in all topologies $\tau _{\alpha }$ ($\alpha \in A$) for arbitrary open $V\subseteq Y$. And this condition translates that $f$ is continuous with respect to all topologies $\tau _{\alpha }$ ($\alpha \in A$) on $X$. $\Box$

Proposition (continuity criterion for the least upper bound topology):

Let $X$ be a topological space, $Y$ a set and $(\tau _{\alpha })_{\alpha \in A}$ a family of topologies on $Y$. Let $f:X\to Y$ be a function. Then $f$ is continuous with respect to the least upper bound topology of the $\tau _{\alpha }$ if and only if $f$ is continuous for all topologies $\tau _{\alpha }$.

Proof: $f$ is continuous iff for all open $V$, the set $f^{-1}(V)$ is open. If that is the case, then all the sets $f^{-1}(V)$ are open, where $V\in \tau _{\alpha }$ ($\alpha \in A$ arbitrary), so that $f$ is continuous with respect to all topologies $\tau _{\alpha }$. Suppose now that $f$ is continuous with respect to all these topologies. Note that since the least upper bound topology on $Y$ with respect to the $\tau _{\alpha }$ is the topology generated by $\bigcup _{\alpha \in A}\tau _{\alpha }$, the latter forms a subbasis of the least upper bound topology. Hence, we may apply the characterisation of continuity via subbasis. $\Box$