Proof: We claim that we have a bijection between sets ${\displaystyle \kappa \subseteq {\mathcal {P}}(X)}$ that satisfy 1.-3. in the proposition (call this set ${\displaystyle {\mathcal {K}}(X)}$) and the topologies on ${\displaystyle X}$, which we shall denote by ${\displaystyle {\mathcal {T}}(X)}$, so that if this bijection is denoted by ${\displaystyle \Phi :{\mathcal {T}}(X)\to {\mathcal {K}}(X)}$, then the closed sets of a topology ${\displaystyle \tau }$ are given precisely by ${\displaystyle \Phi (\tau )}$. Indeed, this map ${\displaystyle \Phi }$ is given by

${\displaystyle \Phi (\tau ):=\{X\setminus U|U\in \tau \}}$, and the inverse is ${\displaystyle \Phi ^{-1}(\kappa )=\{X\setminus F|F\in \kappa }$.

These are inverse to each other by a straightforward computation, and the closed sets of ${\displaystyle \tau }$ are precisely ${\displaystyle \Phi (\tau )}$. We must show that ${\displaystyle \Phi }$ is well-defined, that is, maps ${\displaystyle {\mathcal {T}}(X)}$ to ${\displaystyle {\mathcal {K}}(X)}$ and the inverse is also well-defined in that it maps ${\displaystyle {\mathcal {K}}(X)}$ to ${\displaystyle {\mathcal {T}}(X)}$. Let first ${\displaystyle \tau }$ be a topology. Since ${\displaystyle X\setminus \emptyset =X}$ and ${\displaystyle X\setminus X=\emptyset }$, 1. of the above is satisfied. Suppose now that ${\displaystyle F_{1},\ldots ,F_{n}\in \Phi (\tau )}$, so that for ${\displaystyle j\in [n]}$ we have ${\displaystyle F_{j}=X\setminus U_{j}}$ for a certain ${\displaystyle U_{j}\in \tau }$. Since ${\displaystyle \tau }$ is a topology, ${\displaystyle U_{1}\cap \cdots \cap U_{n}\in \tau }$, and further, by de Morgan

${\displaystyle \Phi (U_{1}\cap \cdots \cap U_{n})=F_{1}\cup \cdots \cup F_{n}}$.

Finally, suppose that ${\displaystyle (F_{\alpha })_{\alpha \in A}}$ is a family of elements of ${\displaystyle \Phi (\tau )}$, and for ${\displaystyle \alpha \in A}$ write again ${\displaystyle F_{\alpha }=X\setminus U_{\alpha }}$. Since ${\displaystyle \tau }$ is a topology, ${\displaystyle \cup _{\alpha \in A}U_{\alpha }\in \tau }$, and by de Morgan

${\displaystyle \Phi \left(\bigcup _{\alpha \in A}U_{\alpha }\right)=\bigcap _{\alpha \in A}F_{\alpha }}$,

and ${\displaystyle \kappa :=\Phi (\tau )}$ satisfies 1. - 3. The reverse direction is proved by analogous computations. Thus, the unique topology whose closed sets are ${\displaystyle \kappa }$ is ${\displaystyle \Phi ^{-1}(\kappa )}$, and if ${\displaystyle \tau }$ is a topology, ${\displaystyle \Phi (\tau )}$, its closed sets, satisfy 1.-3. ${\displaystyle \Box }$

Proof: This is a special case of the general fact that subsets of the power set are ordered by inclusion. ${\displaystyle \Box }$

Proof: This follows from intersections preserving closure properties, by noting that finite intersection and arbitrary union are operations on ${\displaystyle {\mathcal {P}}(X)}$ (as well as are the whole and the empty set, namely 0-ary operations), and that each topology is closed under these. ${\displaystyle \Box }$

Proof: Let's first show that whenever the sets ${\displaystyle N(x)}$ are the neighbourhoods of a topology ${\displaystyle \tau }$ on ${\displaystyle X}$, then they satisfy 1. - 4. Indeed, ${\displaystyle X}$ is an (open) neighbourhood of every of its points. Then, whenever ${\displaystyle U,V\in N(x)}$ for some ${\displaystyle x\in X}$, we find open sets ${\displaystyle U',V'}$ such that ${\displaystyle x\in U'\subseteq U}$ and ${\displaystyle x\in V'\subseteq V}$, note that ${\displaystyle x\in U'\cap V'\subseteq U\cap V}$ and that ${\displaystyle U'\cap V'}$ is open, as finite intersections of open sets are open. If ${\displaystyle S\in N(x)}$, ${\displaystyle x\in U\subseteq S}$ open and ${\displaystyle T\supseteq S}$ arbitrary, then ${\displaystyle x\in U\subseteq T}$ so that ${\displaystyle T\in N(x)}$. Finally, if ${\displaystyle S\in N(x)}$, choose an open ${\displaystyle U}$ s.t. ${\displaystyle x\in U\subseteq S}$, then ${\displaystyle U}$ is open, hence a neighbourhood of all of its points. Now suppose that for each ${\displaystyle x\in X}$ we are given ${\displaystyle N(x)}$ and these neighbourhoods satisfy the conditions 1.-4. Then we define

Failed to parse (unknown function "\middle"): {\displaystyle \tau := \left\{U \subseteq X \middle| \forall x \in U : \exists V \in N(x): V \subseteq U\right\}}

and claim that ${\displaystyle \tau }$ is a topology. Indeed, ${\displaystyle \emptyset \in \tau }$, since then the condition is trivially satisfied, since there are no ${\displaystyle x\in \emptyset }$. Furthermore, ${\displaystyle X\in \tau }$, since whenever ${\displaystyle x\in X}$, ${\displaystyle X\subseteq X}$ and ${\displaystyle X}$ is a neighbourhood of ${\displaystyle x}$. Then, suppose that ${\displaystyle (U_{\alpha })_{\alpha \in A}}$ is a family of sets contained in ${\displaystyle \tau }$, and set ${\displaystyle U:=\cup _{\alpha \in A}U_{\alpha }}$. We claim that ${\displaystyle U\in \tau }$. Indeed, if ${\displaystyle x\in U}$, pick ${\displaystyle \alpha \in A}$ so that ${\displaystyle x\in U_{\alpha }}$, then ${\displaystyle V\in N(x)}$ such that ${\displaystyle x\in V\subseteq U_{\alpha }}$ since ${\displaystyle U_{\alpha }\in \tau }$, and finally note that ${\displaystyle x\in V\subseteq U}$. Then, let ${\displaystyle U,V\in \tau }$ and pick ${\displaystyle U',V'\in N(x)}$ so that ${\displaystyle x\in U'\subseteq U}$ and ${\displaystyle x\in V'\subseteq V}$. Then ${\displaystyle x\in U'\cap V'\subseteq U\cap V}$, so that ${\displaystyle U\cap V\in \tau }$, since ${\displaystyle U'\cap V'\in N(x)}$.

Now we claim that for each ${\displaystyle x\in X}$, the neighbourhoods of ${\displaystyle x}$ with respect to ${\displaystyle \tau }$ are precisely ${\displaystyle N(x)}$. Indeed, let first a neighbourhood ${\displaystyle S}$ of ${\displaystyle x}$ be given. Choose ${\displaystyle x\in U\subseteq S}$ open. Then by definition of ${\displaystyle \tau }$, there exists ${\displaystyle T\in N(x)}$ such that ${\displaystyle T\subseteq U}$, and since ${\displaystyle N(x)}$ is closed under supersets, ${\displaystyle S\in N(x)}$. Conversely, suppose ${\displaystyle S\in N(x)}$. Define ${\displaystyle U:=\{y\in X|S\in N(y)\}}$, and claim that ${\displaystyle U\subseteq S}$ and ${\displaystyle U}$ is open in ${\displaystyle \tau }$. Indeed, ${\displaystyle U\subseteq S}$ is clear, since ${\displaystyle y\in S}$ is necessary for ${\displaystyle S}$ being a neighbourhood of ${\displaystyle y}$. Let now ${\displaystyle y\in U}$. By 4., choose ${\displaystyle U'\subseteq S}$ so that ${\displaystyle U'\in N(y)}$ and ${\displaystyle S}$ is in ${\displaystyle N(z)}$ for all ${\displaystyle z\in U'}$. Thus ${\displaystyle S\in N(z)}$ for all ${\displaystyle z\in U'}$, so that ${\displaystyle U'\subseteq U}$, and since ${\displaystyle y}$ was arbitrary, ${\displaystyle U}$ is open, and ${\displaystyle S}$ is a neighbourhood in ${\displaystyle \tau }$ of ${\displaystyle x}$.

Finally let ${\displaystyle \tau '}$ be any topology which has the sets ${\displaystyle N(x)}$ as neighbourhoods, and claim that ${\displaystyle \tau '=\tau }$ as defined above. Indeed, if ${\displaystyle U\in \tau '}$, then ${\displaystyle U}$ is itself a neighbourhood of each of its points and hence in ${\displaystyle \tau }$. Conversely, if ${\displaystyle U\in \tau '}$, then for each ${\displaystyle x\in U}$ choose a neighourhood ${\displaystyle V_{x}\in \tau }$ of ${\displaystyle X}$ that is contained within ${\displaystyle U}$ (in order to avoid the axiom of choice, choose them to be the union of all such sets) and note that

${\displaystyle U=\bigcup _{x\in X}V_{x}\in \tau }$. ${\displaystyle \Box }$

Exercises

1. Let ${\displaystyle X}$ be a set.
   1. Prove that ${\displaystyle {\mathcal {P}}(X)}$, the power set, is a topology on ${\displaystyle X}$ (it's called the discrete topology) and that when ${\displaystyle X}$ is equipped with this topology and ${\displaystyle f:X\to Y}$ is any function where ${\displaystyle Y}$ is a topological space, then ${\displaystyle f}$ is automatically continuous.
   2. Prove that ${\displaystyle \{\emptyset ,X\}}$ is a topology on ${\displaystyle X}$ (called the trivial topology), and that when ${\displaystyle X}$ is equipped with this topology, ${\displaystyle Z}$ is any topological space and ${\displaystyle f:Z\to X}$ is any function, then ${\displaystyle f}$ is continuous.