# General Topology/Constructions

To every set, we may associate its boundary, its interior and its closure.
Definition (boundary):
Let $X$ be a topological space and $S\subseteq X$ a subset. The boundary of $S$ is defined as the set
    $\partial S:=\left\{x\in X\mid \forall M\in N(X):M\cap S\neq \emptyset \wedge M\cap (X\setminus S)\neq \emptyset \right\}$.
Proposition (a set is closed iff it contains its boundary):
Let $X$ be a topological space. A set $A\subseteq X$ is closed if and only if $\partial A\subseteq A$.
Proof: Suppose first that $A$ contains its boundary. We show that $X\setminus A$ is open. Indeed, let $y\in X\setminus A$. Note that a set is open if and only if it contains a neighbourhood of each of its points. If there does not exist a neighbourhood $M\subseteq X$ so that $y\in M$ and $M\cap A=\emptyset$, then since $y\in X\setminus A$, we get $y\in \partial A$, a contradiction. Suppose then that $A$ is closed, and let $y\notin A$. If there does not exist a neighbourhood $M$ of $y$ contained in $X\setminus A$, then $y\in \partial A\subseteq A$, contradiction. $\Box$
Definition (interior):
Let $X$ be a topological space and $A\subseteq X$ be a set. The interior of $A$ is defined to be
    ${\overset {\circ }{A}}:=A\setminus \partial A$.
Proposition (the interior of a set is open):
Let $X$ be a topological space and $A\subseteq X$. Then ${\overset {\circ }{A}}$ is open.
Proof: Let $y\in {\overset {\circ }{A}}$. Then there exists a neighbourhood of $y$ contained in ${\overset {\circ }{A}}$, for otherwise, all neighbourhoods $M$ of $y$ intersect $A$ (since $y\in A$) as well as $X\setminus A$, since if $M$ is a neighbourhood that does not intersect $X\setminus A$, but that does intersect $\partial A$ we may pass to an open subset $U$ of $M$ which is still a neighbourhood of $y$ and intersects $\partial A$ (otherwise we are done), so that $x\in U\cap \partial A$, so that $U$ is a neighbourhood of $x\in \partial A$, so that $U$ intersects $X\setminus A$. Hence ${\overset {\circ }{A}}$ is open by the characterisation of topologies by neighbourhood systems. $\Box$
Definition (closure):
Let $X$ be a topological space and let $A\subseteq X$. The closure of $A$ is defined to be
    ${\overline {A}}:=A\cup \partial A$.
Proposition (the closure of a set is closed):
Let $X$ be a topological space and $A\subseteq X$ a set. Then ${\overline {A}}$ is closed.
Proof: We show that $X\setminus {\overline {A}}$ is open. Let $x\notin {\overline {A}}$. Suppose that $X\setminus {\overline {A}}$ does not contain a neighbourhood of $x$. Then every neighbourhood of $x$ intersects either $A$ or $\partial A$. If $M$ is a neighbourhood of $x$ that intersects $\partial A$ but not $A$, we may pick $M$ to be open by passing to a subset of $M$, and then we'll still have some point $y\in \partial A\cap M$, since otherwise the original $M$ would intersect $A$. But $y\in \partial A$ implies by definition that all neighbourhoods of $y$ (e.g. $M$...) intersect $A$, so that we obtain a contradiction, so that all neighbourhoods of $x$ intersect $A$ and $x\in \partial A$, contradiction. $\Box$
Proposition (the closure of a set equals the intersection of all closed subsets contained within it):
Let $X$ be a topological space and $A\subseteq X$ a set. Then
    ${\overline {A}}=\bigcap _{B{\text{ closed}} \atop B\supseteq A}B$.
Proof: We have seen that the closure of a set is closed, so that the right hand side is contained in the left hand side. Further, let $x\in {\overline {A}}$ and let $B$ be a closed superset of $A$. Suppose that $x\notin B$, then there exists $M\in N(x)$ so that $M\cap B=\emptyset$ since $X\setminus B$ is open. But then neither $x\in A$ nor $x\in \partial A$, since $M\cap B\supseteq M\cap A$, a contradiction. $\Box$
Definition (dense):
Let $X$ be a topological space and $S\subseteq X$ a subset. $S$ is called dense in $X$ iff ${\overline {S}}=X$.
Proposition (characterisation of density):
Let $X$ be a topological space and $S\subseteq X$. The following are equivalent:
* $S$ is dense in $X$
* for all nonempty open subsets $\emptyset \subsetneq U\subseteq X$, we have $S\cap U\neq \emptyset$
* the interior of $X\setminus S$ is empty
Proof: The equivalence of 1. and 3. follows because of the fact that $X\setminus {\overline {S}}={\overset {\circ }{X\setminus S}}$, which in turn can be seen from the definition of interior and closure and the fact that $\partial S=\partial (X\setminus S)$ in general. Now we prove the equivalence of 1. and 2. Indeed, suppose first that $S$ is dense in $X$, and let $U\subseteq X$ be open and nonempty. If $U\cap S=\emptyset$, then $X\setminus U$ is a proper closed superset of $S$, whence ${\overline {S}}\subseteq X\setminus U$, a contradiction. Suppose now that 2. holds, and let $A\supseteq S$ be any closed superset of $S$. Then suppose that $x\notin A$ for some $x\in X$. Then $X\setminus A$ is a nonempty open set that does not intersect $S$, contradicting 2. Hence, $A=X$, and we conclude since the closure of a set equals the intersection of all closed subsets contained within it. $\Box$

## Upper and lower bounds

Definition (greatest lower bound topology):
Let $X$ be a set and let $(\tau _{\alpha })_{\alpha \in A}$ be a family of topologies on $X$. The topology given by
    $\bigcap _{\alpha \in A}\tau _{\alpha }$,
being the largest topology that is contained in all the $\tau _{\alpha }$, is called the greatest lower bound topology
Definition (topology generated by a set):
Let $X$ be a set and let ${\mathcal {B}}\subseteq {\mathcal {P}}(X)$. The topology generated by ${\mathcal {B}}$ is the topology given by
    $\bigcap _{\tau {\text{ topology on }}X \atop {\mathcal {B}}\subseteq \tau }\tau$.
Note that these two are topologies since the intersection of topologies is again a topology.
Proposition (characterisation of the generated topology):
Let $X$ be a set and ${\mathcal {B}}\subseteq {\mathcal {P}}(X)$. Let $\tau$ be the topology generated by ${\mathcal {B}}$. Then, upon defining
    $\sigma :=\left\{\bigcup _{\alpha \in A}C_{\alpha }\mid A{\text{ index set }},C_{\alpha }=B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }},\forall j\in [n_{\alpha }]:B_{\alpha ,n}\subseteq {\mathcal {B}}\right\}$,
    $\sigma =\tau$.
Proof: Clearly, $\sigma \subseteq \tau$, since $\tau$ is a topology that contains all the $B_{\alpha ,j}$. On the other hand, we show that $\sigma$ is a topology, thereby proving that $\tau \subseteq \sigma$ by definition of the topology generated by a set. Indeed, note first that $X\in \sigma$ upon choosing $A=\{1\}$ and $C_{1}:=X$, the empty intersection. Then note that $\emptyset \in \sigma$ by choosing $A=\emptyset$. Suppose then that $U,V\in \sigma$ and write, for $j\in [n]$,
    $U=\bigcup _{\alpha \in A}C_{\alpha }$, $C_{\alpha }=B_{\alpha ,1}\cap \cdots \cap B_{\alpha ,n_{\alpha }}$ and $V=\bigcup _{\omega \in \Omega }D_{\omega }$, $D_{\omega }=B_{\omega ,1}\cap \cdots \cap B_{\omega ,n_{\omega }}$.
Then
    $U\cap V=\bigcup _{\alpha \in A}\bigcup _{\omega \in \Omega }C_{\alpha }\cap D_{\omega }\in \sigma$.
Finally, suppose that for $i\in I$ we have $U_{i}\in \sigma$, and write
    $U_{i}:=\bigcup _{\alpha \in A_{i}}C_{\alpha }$, $C_{\alpha }$ being a finite intersection of sets of ${\mathcal {B}}$.
Then
    $\bigcup _{i\in I}\bigcup _{\alpha \in A_{i}}C_{\alpha }=\bigcup _{\alpha \in \cup _{i}A_{i}}C_{\alpha }\in \sigma$.
Thus, we've shown that $\sigma$ is a topology. $\Box$
Definition (least upper bound topology):
Let $X$ be a set, and let $(\tau _{\alpha })_{\alpha \in A}$ be topologies on $X$. The least upper bound topology on $X$ is defined to be the topology generated by
    $\bigcup _{\alpha \in A}\tau _{\alpha }$.
Proposition (least upper bound topology is smallest topology containing the given topologies):
Let $X$ be a set, and let $(\tau _{\alpha })_{\alpha \in A}$ be topologies on $X$. Let $\tau$ be the least upper bound topology on $X$ defined by the $\tau _{\alpha }$. Then $\tau$ is minimal among all topologies that contain all $\tau _{\alpha }$ ($\alpha \in A$).
Proof: By definition of the topology generated by a set, any topology $\tau '$ that contains all the topologies $\tau _{\alpha }$ will contain $\tau$. $\Box$
Proposition (topologies on a fixed set are a complete lattice):
Let $X$ be a set, and let $(\tau _{i})_{i\in I}$ be a family of topologies on $X$. Then upon ordering $X$ by inclusion, there exists a least upper bound of $(\tau _{i})_{i\in I}$, as well as a greatest lower bound. Thus, the topologies on a set, when ordered by inclusion, form a complete lattice.
Proof: We have already seen that there exists a least upper bound topology. Then note that topologies form an algebraic variety, where union and intersection are arbitr-ary operations, whereas the whole set and the empty set are $0$-ary operations. Then the existence of the greatest lower bound (and that it equals the intersection of the topologies in question) follows immediately since the greatest lower bound structure is the intersection. $\Box$

## Final and initial topologies

Proposition (preimage of a topology is a topology):
Let $f:X\to Y$ be a function from a set $X$ to another set $Y$, and let $\tau$ be a topology on $Y$. Then $f^{-1}(\tau )$ is a topology on $X$.
Proof: This follows instantly from the formulae
* $f^{-1}(Y)=X$, $f^{-1}(\emptyset )=\emptyset$
* $f^{-1}\left(\bigcup _{i\in I}B_{i}\right)=\bigcup _{i\in I}f^{-1}(B_{i})$ and $f^{-1}\left(\bigcap _{i\in I}B_{i}\right)=\bigcap _{i\in I}f^{-1}(B_{i})$ for any family of subsets $(B_{i})_{i\in I}$ of $Y$ $\Box$
Definition (initial topology):
Let $X$ be a set and let $(X_{i},\tau _{i})_{i\in I}$ be a family of topological spaces. Let further $f_{i}:X\to X_{i}$ ($i\in I$) be functions. The initial topology on $X$ is the least topology that contains all the topologies $f_{i}^{-1}(\tau _{i})$.
Definition (product topology):
Let $(X_{\alpha })_{\alpha \in A}$ be a family of topological spaces. Define the set
    $X:=\prod _{\alpha \in A}X_{\alpha }$, the Cartesian product.
The product topology is a topology on $X$, namely the initial topology with respect to the collection of all projections
    $\pi _{\alpha }:X\to X_{\alpha },\pi _{\alpha }((x_{\beta })_{\beta \in A})=x_{\alpha }$, $\alpha \in A$.