\chapter{Constructions of topological spaces}


Now that we are familiar with the notion of a topological space and what they represent, it is useful to look at some general methods that can be used to construct new topological spaces from existing ones.

\section{Topological subspaces}

To extend a topology to a subset of a topological space, we introduce topological subspaces.

\begin{definition}
A topological subspace of $(X,\mathcal{T})$ is a topological space $(Y,\mathcal{T}_{Y})$.
\begin{itemize}
\item $Y \subseteq X$ is a subset of $X$
\item $\mathcal{T}_Y = \{ Y \cap U : U \in \mathcal{T}\}$ is the induced topology for the topological subspace 
\end{itemize}
\end{definition}

Indeed, topological subspaces always form a topological space, however its topological properties aren't necessarily (and often aren't) the same as the original space. That said, much can be said regarding a topological subspaces' relationship with the original space.

\begin{proposition}
Let $Y$ be a topological subspace of $X$. A set if closed in $Y$ iff it is of the form $Y \cap F$ where $F$ is closed in $X$.
\end{proposition}


\begin{proposition}
Let $Y$ be a topological subspace of $X$. If $\mathcal{B}$ is a basis for $\mathcal{T}$, then $\mathcal{B}_Y = \{ B \cap Y : B \in \mathcal{B} \}$ is a basis for $\mathcal{T}_{Y}$
\end{proposition}


\begin{proposition}
Let $Y$ be a topological subspace of $X$. If $Y$ is open in $X$, then open sets in $Y$ are open sets in $X$.
Let $Y$ be a topological subspace of $X$. If $Y$ is closed in $X$, then closed sets in $Y$ are closed sets in $X$.
\end{proposition}


\begin{proposition}
Let $Y$ be a topological subspace of $X$ and $A \subseteq Y$/
\[\mathrm{cl}_{X}(A) \cap Y = \mathrm{cl}_{Y}(A)\]
\end{proposition}





\section{Product topological spaces}

\begin{definition}[Box topological space]
Let $(X_i , \mathcal{T}_i)_{i \in I}$ be a family of topological spaces. The \emph{box topological space} $\prod_{j \in I}X_j$ is the topological space generated by the following basis.
	\[\mathcal{B} = \{ \prod{j \in I} U_j : U_j \in \mathcal{T}_j \}\]
Essentially, the cartesian product of open sets from each $X_j$ is a basis element.
\end{definition}


This is not the categorical product in the $\mathrm{Top}$ category, since there is a counterexample for which there doesn't exist a continuous function that satisfies the property.




The ideal product topology would be defined with its continuous \emph{projection maps} $\pi_i (x_1,x_2,\hdots) = x_i$ and the property that  that for a family of continuous $f_i : Y \to X_i$, there is some unique and continuous $f : Y \to \prod X_i$ where $\pi_i \circ f = f_i$.

Although all the projection maps are continuous for the box topology, unfortunately it is possible for the unique $f$ to fail to be continuous. The reason for this is actually rooted in the fact that open sets are only closed under finite intersections;

In search of a better construction, we consider the sets X_1xX_2x...xU_ix... (where U_i is open in X_i); these sets must be open in our desired $\prod X_i$, since $f^{-1}(X_1xX_2x...xU_ix...) = f_{i}^{-1}(U_i)$ is open in $Y$ due to the continuity of $f_i : Y \to X_i$, and so if $f$ is to be continuous, this cartesian product would have to be open. Notice that any open set inthe box topology is a (possibly infinite) intersection of such sets, and the preimage of this infinite intersection (using that $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$ generally, and again that $f^{-1}(X_1xX_2x...xU_ix...) =f_{i}^{-1}(U_i)$ are open sets of $Y$ by continuity of $f_i$) is $\bigcap f_i^{-1}(U_i)$, which is only guaranteed to be open in $Y$ when the intersection is finite.

If one only allows finite intersections of this form, then this forms a basis for a topology that satisfies our ideal property. 

\begin{definition}[Product topological space]
For each $i \in \mathbb{N} \cap[1,n]$, let $(X_i , \mathcal{T}_i)$ be topological spaces. The \emph{box topological space} $\prod_{j \in I}X_j$ is the topological space generated by the following basis.
Let $(\prod^{n}_{i=1} X_i, \mathcal{T})$ be a box topological space, then all $\mathcal{T}$ is the coarsest topology such that all projection functions are continuous.
\end{definition}

This product topology means that continuous arguments (i.e the $f_i$) is enough for a continuous $f$ (as our ideal property alludes to).

also note the following.

\begin{proposition}
product topology coarsest topology permitting continuous projections
\end{proposition}

Why is this so? If a set of the form X_1xX_2x...xU_ix... is forcibly removed from the topology, then $ \pi_{i}^{-1}(U_i) = X_1xX_2x...xU_ix...$ is not open in $\prod X_i$ even though U_i is open in X_i, meaning that $\pi_i$ fails to be continuous. If some X_1xX_2x...xU_ix...xU_jx... is removed, then the conditions for a topology won't hold since it won't be closed under finite intersections.


finite products mean product topology equals box topology


\begin{proposition}
Let $(\prod^{n}_{i=1} X_i, \mathcal{T})$ be a box topological space, then all $\mathcal{T}$ is the coarsest topology such that all projection functions are continuous.
\end{proposition}

This result doesn't generalize for infinite box topological spaces.




When we consider infinite products, the projection functions may not be continuous.
We define product topological spaces to be more robust.



\begin{definition}[Box topological space]
projection functions are all continuous.
\end{definition}


\section{Quotient topological spaces}



How can we formally define the idea of constructing a new topological space  'gluing' points of the original space?

The idea is to use equivalence relations as our 'glue'; we want points within the same equivalence class to be seen as the same point (i.e glued together)

Quotient spaces do exactly this; the term 'quotient' appears because by using equivalence clases, we partition (i.e 'divide') our original set.

\begin{definition}[Quotient space]
Given a topological space $(X,\mathcal{T})$ and equivalence relation $\sim$ on $X$, let $X / \sim$ be the set of equivalence classes and $q : x \to X /\sim$ be a surjective function defined as $q(x) = [x]$, called the \emph{quotient map}.

The \emph{quotient topology} $\mathcal{T}_{X / \sim}$ is the topology  $\mathcal{T}_{X / \sim} = \{ V \in X / \sim : q^{-1}(V) \in \mathcal{T}\}$. The topological space $(X / \sim , \mathcal{T}_{X / \sim})$ is the \emph{quotient space induced by $\sim$}.
\end{definition}





\section{Final topology}

We can generalize the idea of our quotient space 

\begin{definition}[Final topology]
Given a set $X$ and topological spaces $(Y_i,\mathcal{U}_i)$ with functions $f_i : Y_i \to X$, the \emph{final topology on $X$} is the finest topology $\mathcal{T}$ on $X$ such that all $f_i$ are continuous.
\end{definition}