Is Projection Function Continuous Unde Rbox Topology

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. ${\displaystyle \Box }$

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. ${\displaystyle \Box }$

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. ${\displaystyle \Box }$

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.

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. ${\displaystyle \Box }$

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. ${\displaystyle \Box }$

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 ${\displaystyle S}$ . ${\displaystyle \Box }$

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. ${\displaystyle \Box }$

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.

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 ${\displaystyle f(X)}$ .

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 ${\displaystyle X}$ . ${\displaystyle \Box }$

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. ${\displaystyle \Box }$

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Source: https://en.wikibooks.org/wiki/General_Topology/Continuity

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