# General Topology/Continuity

Definition (continuity):

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

Proposition (characterisation of continuity via subbasis):

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

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

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

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

Example (identity is a continuous function):

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

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

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

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

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

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

Proposition (product of topological spaces is categorical product):

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

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

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

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

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

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

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

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

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

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

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

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

Proposition (characterisation of continuity by closed sets):

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

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

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

Proposition (continuity on closed union):

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

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

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

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

Proposition (restriction of continuous function is continuous):

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

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

Definition (homeomorphism):

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

Proposition (restriction of homeomorphism is homeomorphism):

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

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

Definition (equicontinuity):

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

Proposition (function in equicontinuous set is continuous):

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

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

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

Definition (local homeomorphism):

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

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

Proposition (local homeomorphisms are open):

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

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

Definition (embedding):

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

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

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

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

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

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

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