General Topology/Definition, characterisations

Definition (topology):

Let $X$ be any set. A topology on $X$ is a subset $\tau \subset {\mathcal {P}}(X)$ of the power set of $X$ such that the following three axioms hold:

1. $X\in \tau$ and $\emptyset \in \tau$
2. $U_{1},\ldots ,U_{n}\in \tau \Rightarrow U_{1}\cap \cdots \cap U_{n}\in \tau$
3. $(\forall i\in I:U_{i}\in \tau )\Rightarrow \bigcup _{i\in I}U_{i}\in \tau$

Definition (topological space):

A topological space is a set $X$ together with a topology $\tau$ on it.

Example (Euclidean topology):

Let $n\in \mathbb {N}$ and consider the set $X=\mathbb {R} ^{n}$. Then we define

$\tau :=\{U\subseteq \mathbb {R} ^{n}\mid \forall x\in U:\exists \epsilon >0:B_{\epsilon }(x)\subseteq U\}$.

$\tau$ is a topology on $\mathbb {R} ^{n}$, called the Euclidean topology.

Definition (open set):

Let $X$ be a topological space and $\tau$ its topology. An open set of $X$ is any set contained in $\tau$.

Definition (closed set):

Let $X$ be a topological space. A subset $A\subset X$ is called closed if and only if $X\setminus A$ is open, ie. contained in the topology of $X$.

Definition (clopen set):

Let $X$ be a topological space. A subset $A\subset X$ is called clopen if and only if it is both open and closed at the same time.

Proposition (definition of topologies from closed sets):

Let $X$ be a set, and let $\kappa \subseteq {\mathcal {P}}(X)$ be a set of subsets of $X$ such that the following axioms hold:

1. $\emptyset \in \kappa$ and $X\in \kappa$
2. $F_{1},\ldots ,F_{n}\in \kappa \Rightarrow F_{1}\cup \cdots \cup F_{n}\in \kappa$
3. $(\forall \alpha \in A:F_{\alpha }\in \kappa )\Rightarrow \bigcap _{\alpha \in A}F_{\alpha }\in \kappa$

Then there exists a unique topology $\tau$ on $X$ such that $\kappa$ contains precisely the closed sets of that topology, and conversely the closed sets of any topology satisfy 1. - 3.

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

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

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

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

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

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

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

Proposition (topologies are ordered by inclusion):

If $X$ is any set and $\tau _{1},\tau _{2}$ are topologies on $X$, then $\tau _{1}\leq \tau _{2}:\Leftrightarrow \tau _{1}\subseteq \tau _{2}$ defines an order on the topologies of $X$.

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

Definition (comparison of topologies):

Let $X$ be any set and $\tau _{1},\tau _{2}$ topologies on $X$. Whenever $\tau _{1}\subseteq \tau _{2}$, we say that $\tau _{1}$ is coarser than $\tau _{2}$ and $\tau _{2}$ is finer than $\tau _{1}$.

Proposition (intersection of topologies is a topology):

Let $(\tau _{\alpha })_{\alpha \in A}$ be a family of topologies on a set $X$. Then

$\bigcap _{\alpha \in A}\tau _{\alpha }$

is a topology on $X$.

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

Definition (open neighbourhood):

Let $(X,\tau )$ be a topological space, and let $x\in X$. A neighbourhood of $x$ is an open set $U$ (ie. $U\in \tau$) such that $x\in U$.

Definition (neighbourhood):

Let $(X,\tau )$ be a topological space, and let $x\in X$ A neighbourhood of $x$ is a set $N\subset X$ such that there exists an open set $U\in \tau$ such that $U\subseteq N$ and $x\in U$. The set of all neighbourhoods of $x$ is denoted $N(x)$.

Theorem (characterisation of topologies by neighbourhood systems):

Suppose that $X$ is a set, and that for each $x\in X$ we are given a set $N(x)\subseteq {\mathcal {P}}(X)$ such that $x\in S$ for each $S\in N(x)$. Then the sets $N(x)$ form the neighbourhoods of a topology on $X$ if and only if the following conditions are satisfied:

1. $\forall x\in X:X\in N(x)$
2. $\forall x\in X:\forall U,V\in N(x):U\cap V\in N(x)$
3. $\forall x\in X:\forall U\in N(x):\forall S\subseteq X:(U\subseteq S\Rightarrow S\in N(x))$ (closedness under supersets)
4. $\forall x\in X:\forall S\in N(x):\exists U\subseteq S:U\in N(x)\wedge \forall y\in U:S\in N(y)$

In this case, the topology of which the $N(x)$ are the neighbourhoods is uniquely determined by the neighbourhoods and equals

$\left\{U\subseteq X\mid \forall x\in U:\exists V\in N(x):V\subseteq U\right\}$.

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

$\tau :=\left\{U\subseteq X\mid \forall x\in U:\exists V\in N(x):V\subseteq U\right\}$

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

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

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

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

Exercises

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