Equivalent definition of Hausdorff space via Hom and its applications

Equivalent definition of Hausdorff space

Definition. Let $X$$X$ be a topological space. Let $f:Y\to Z$$f: Y \to Z$ be a continuous function such that $\overline{f(Y)}=Z$$\overline{f(Y)} = Z$ (i.e., $f$$f$ is a dense map).
Then $X$$X$ is Hausdorff if the induced map $\mathsf{Top}(f,X):\mathsf{Top}(Z,X)\to \mathsf{Top}(Y,X)$$\mathsf{Top}(f, X): \mathsf{Top}(Z, X) \to \mathsf{Top}(Y, X)$ is injective.

Theorem. The above definition is equivalent to the standard definition of a Hausdorff space.

Proof ($\Rightarrow$$\Rightarrow$). Let $X$$X$ be a Hausdorff space, and let $g,h:Z\to X$$g, h: Z \to X$ be two continuous functions such that $g\circ f=h\circ f$$g \circ f = h \circ f$.
Since $X$$X$ is Hausdorff, the equalizer $\mathsf{Eq}(h,g)=\{z\in Z\mid h(z)=g(z)\}=(h,g{)}^{-1}(\mathrm{\Delta })$$\mathsf{Eq}(h, g) = \{ z \in Z \mid h(z) = g(z) \} = (h, g)^{-1}(\Delta)$ is a closed subset of $Z$$Z$. Since $g$$g$ and $h$$h$ agree on $f(Y)$$f(Y)$, we have $f(Y)\subseteq \mathsf{Eq}(h,g)$$f(Y) \subseteq \mathsf{Eq}(h, g)$. Given that $\overline{f(Y)}=Z$$\overline{f(Y)} = Z$ and the equalizer is closed, it follows that $\mathsf{Eq}(h,g)=Z$$\mathsf{Eq}(h, g) = Z$. Hence, $h=g$$h = g$, proving $\mathsf{Top}(f,X)$$\mathsf{Top}(f, X)$ is injective.

Proof ($\Leftarrow$$\Leftarrow$). Let $X$$X$ be a space such that $\mathsf{Top}(-,X)$$\mathsf{Top}(-, X)$ satisfies the injectivity property for all dense maps. Assume for contradiction that the diagonal $\mathrm{\Delta }\subset X^2$$\Delta \subset X^2$ is not closed.
Consider the dense embedding $j:\mathrm{\Delta }\hookrightarrow \bar{\mathrm{\Delta }}$$j: \Delta \hookrightarrow \overline{\Delta}$. By our assumption, the induced map $j^{\ast }:\mathsf{Top}(\bar{\mathrm{\Delta }},X)\to \mathsf{Top}(\mathrm{\Delta },X)$$j^*: \mathsf{Top}(\overline{\Delta}, X) \to \mathsf{Top}(\Delta, X)$ is injective.
Now consider the two projections ${\pi }_1,{\pi }_2:\bar{\mathrm{\Delta }}\to X$$\pi_1, \pi_2: \overline{\Delta} \to X$. We know that ${\pi }_1\ne {\pi }_2$$\pi_1 \neq \pi_2$ because there exists some point $(a,b)\in \bar{\mathrm{\Delta }}\setminus \mathrm{\Delta }$$(a, b) \in \overline{\Delta} \setminus \Delta$ where $a\ne b$$a \neq b$. However, on the diagonal itself, ${\pi }_1\circ j={\pi }_2\circ j$$\pi_1 \circ j = \pi_2 \circ j$. This contradicts the injectivity of $j^{\ast }$$j^*$. Thus, $\mathrm{\Delta }$$\Delta$ must be closed, and $X$$X$ is Hausdorff. $◻$$\square$

Connectedness

Definition. Let $2:=\{0,1\}$$2 := \{0, 1\}$ be the discrete two-point space (the coproduct $1\coprod 1$$1 \coprod 1$ in $\mathsf{Top}$$\mathsf{Top}$). A topological space $X$$X$ is connected if $|\mathsf{Top}(X,2)|\le 2$$|\mathsf{Top}(X, 2)| \le 2$.

Corollary. Let $U\subseteq Y$$U \subseteq Y$ be a connected subspace. Then for any $Z$$Z$ such that $U\subseteq Z\subseteq \bar{U}$$U \subseteq Z \subseteq \overline{U}$, $Z$$Z$ is connected as well.

Proof. The inclusion map $i:U\hookrightarrow Z$$i: U \hookrightarrow Z$ has a dense image in $Z$$Z$. Since the discrete space $2$$2$ is Hausdorff, the induced map $\mathsf{Top}(Z,2)\hookrightarrow \mathsf{Top}(U,2)$$\mathsf{Top}(Z, 2) \hookrightarrow \mathsf{Top}(U, 2)$ is injective.
Therefore, $|\mathsf{Top}(Z,2)|\le |\mathsf{Top}(U,2)|\le 2$$|\mathsf{Top}(Z, 2)| \le |\mathsf{Top}(U, 2)| \le 2$. Thus, $Z$$Z$ is connected. $◻$$\square$

Unique Limits

Definition. Let $\mathbb{N}$$\mathbb{N}$ be the set of natural numbers with the discrete topology, and let ${\mathbb{N}}^{\ast }=\mathbb{N}\cup \{\mathrm{\infty }\}$$\mathbb{N}^* = \mathbb{N} \cup \{\infty\}$ be its one-point compactification.
Let $i:\mathbb{N}\to {\mathbb{N}}^{\ast }$$i: \mathbb{N} \to \mathbb{N}^*$ be the natural inclusion map. For any topological space $X$$X$:
Sequences in $X$$X$ have unique limits if and only if $\mathsf{Top}(i,X):\mathsf{Top}({\mathbb{N}}^{\ast },X)\to \mathsf{Top}(\mathbb{N},X)$$\mathsf{Top}(i, X): \mathsf{Top}(\mathbb{N}^*, X) \to \mathsf{Top}(\mathbb{N}, X)$ is injective.

Corollary. Every Hausdorff space has unique limits for sequences.

Proof. The inclusion $i:\mathbb{N}\to {\mathbb{N}}^{\ast }$$i: \mathbb{N} \to \mathbb{N}^*$ is a dense map. Since $X$$X$ is Hausdorff, the map $\mathsf{Top}(i,X)$$\mathsf{Top}(i, X)$ is injective by our first theorem. Thus, any sequence (represented by a map from $\mathbb{N}$$\mathbb{N}$) has at most one continuous extension to ${\mathbb{N}}^{\ast }$$\mathbb{N}^*$ (the limit). $◻$$\square$

We know $\mathsf{Top}$$\mathsf{Top}$ is not a topos, however, we have the open/closed subobject classifier, $\mathrm{\Omega }:=(\{\mathrm{\perp },\mathrm{\top }\},\{\{\mathrm{\perp },\mathrm{\top }\},\{\mathrm{\top }\},\mathrm{\varnothing }\})$$\Omega:=(\{\bot,\top\},\{\{\bot,\top\},\{\top\},\emptyset\})$

The only nontrivial closed subset is $\{\mathrm{\perp }\}$$\{\bot\}$. Hence $\mathsf{Top}(X,\mathrm{\Omega })\cong \mathsf{Cl}(X)$$\mathsf{Top}(X,\Omega)\cong\mathsf{Cl}(X)$.

In a topos, the diagonal map gives us the predicate of $=$$=$. That is, consider the subobject ${\mathrm{\Delta }}_X:X\to X\times X$$\Delta_X:X\to X\times X$, it corresponds to a morphism ${\delta }_X:X\times X\to \mathrm{\Omega }$$\delta_X:X\times X\to\Omega$. In $\mathsf{Set}$$\mathsf{Set}$, this is just $\delta (x,y)=\mathrm{\top }⟺x=y$$\delta(x,y)=\top\iff x=y$.

Let $f,g:U\to X$$f,g:U\to X$ be two general elements, if ${\delta }_X\circ (f,g)={\mathrm{\top }}_U:U\to \mathrm{\Omega }$$\delta_X\circ(f,g)=\top_U:U\to \Omega$, then we have $f=g$$f=g$.

Indeed, ${\delta }_X\circ (f,g)$$\delta_{X}\circ(f,g)$ is the ${\chi }_{\mathsf{Eq}(f,g)}$$\chi_{\mathsf{Eq}(f,g)}$. Well, although $\mathsf{Top}$$\mathsf{Top}$ is not a topos, for a Hausdorff space $X$$X$, ${\mathrm{\Delta }}_X:X\to X^2$$\Delta_X:X\to X^2$ is a closed subobject. Hence we could consider ${\delta }_X\circ (f,g)$$\delta_X\circ (f,g)$. Let $V\subseteq U$$V\subseteq U$ be a subset, then ${\delta }_X\circ (f,g)(V)=\{\mathrm{\top }\}⟺V\subseteq \mathsf{Eq}(f,g).$$\delta_X\circ(f,g)(V)=\{\top\}\iff V\subseteq\mathsf{Eq}(f,g).$

As a corollary, the equalizer of $(f,g)$$(f,g)$ is closed.

Also, assume that $f{|}_U=g{|}_U$$f|_U=g|_U$ where $U$$U$ is a dense subset. Then $f=g$$f=g$. Because ${\delta }_X\circ (f,g)(\bar{U})\subseteq \overline{{\delta }_X\circ (f,g)(U)}=\{\mathrm{\top }\}$$\delta_X\circ(f,g)(\overline{U})\subseteq \overline{\delta_X\circ(f,g)(U)}=\{\top\}$

Another Corollary:

Let $G$$G$ be a topological group. Then $G$$G$ is Hausdorff iff $\{e\}$$\{e\}$ is closed.

Proof. $\{e\}$$\{e\}$ is closed iff ${\chi }_{\{e\}}:G\to \mathrm{\Omega }$$\chi_{\{e\}}:G\to\Omega$ is continuous.

Notice that ${\delta }_G(x,y)=\mathrm{\top }⟺x=y⟺x{y}^{-1}=e$$\delta_G(x,y)=\top\iff x=y\iff xy^{-1}=e$, hence ${\delta }_G={\chi }_{\{e\}}\circ m\circ ({\mathrm{id}}_G,\iota )$$\delta_G=\chi_{\{e\}}\circ m\circ (\mathrm{id}_G,\iota)$.

Here $\iota :x⟼x^{-1}$$\iota:x\longmapsto x^{-1}$. Then if ${\chi }_{\{e\}}$$\chi_{\{e\}}$ is continuous, ${\delta }_G$$\delta_G$ is continuous.

If you do not know that in Hausdorff spaces, the singleton $\{e\}$$\{e\}$ is closed, then consider that if ${\delta }_G$$\delta_G$ is continuous, then ${\delta }_G(x,e)={\chi }_{\{e\}}(x)$$\delta_G(x,e)=\chi_{\{e\}}(x)$ is continuous. $◻$$\square$

Let $\mathrm{\Gamma }=\{(x,f(x))\in X\times Y:x\in X\}$$\Gamma=\{(x,f(x))\in X\times Y: x\in X\}$. If $Y$$Y$ is a Hausdorff space, then $\mathrm{\Gamma }$$\Gamma$ is closed.

Proof.

${\chi }_{\mathrm{\Gamma }}={\delta }_Y\circ (f\times {\mathrm{id}}_Y)$$\chi_\Gamma = \delta_{Y}\circ (f\times \mathrm{id}_Y)$ is a composition of continuous functions, hence continuous. Therefore, $\mathrm{\Gamma }$$\Gamma$ is a closed subspace.