# A space is $T_0$ if and only if it is homeomorphic to a subpspace of $S^I$ with $S$ the Sierpinski space for some $I$ [Proof Verification]

As the title states, I want to check my proof of the following:

Proposition. A topological space $$(X, \tau)$$ is $$T_0$$ if and only if there exists a set $$I$$ such that $$(X,\tau)$$ is homeomorphic to a subpspace of $$S^I$$ with $$S = (\{0,1\}, \{\emptyset, \{1\}, \{0,1\}\})$$ the Sierpinski space.

One direction is straightforward: since $$S^I$$ is a product of $$T_0$$, it is itself $$T_0$$ and so is every subspace. In particular, via the homeomorphism of $$X$$ with the subspace of $$S^I$$, we get that $$X$$ is $$T_0$$.

The other implication is the one I am having doubts about.

Suppose that $$X \neq \emptyset$$ (otherwise take $$I = \emptyset$$) and let $$\mathcal{C} = \{f : X \to S : \text{f is continuous} \}$$ which is not empty, for example because $$X \to \{1\} \hookrightarrow S$$ is continuous. Now, consider

\begin{align*} \Gamma : & \ X \longrightarrow \prod_{f \in \mathcal{C}}S \\ & x \longmapsto (f(x))_{f \in \mathcal{C}} \end{align*}

which is continuous because each $$\pi_f \Gamma \equiv f$$ is continuous. This function is injective: if $$x \neq y$$, since $$X$$ is $$T_0$$ without loss of generality there exists an open set $$y \in U$$ of $$X$$ with $$x \not \in U$$. Thus,

$$\Gamma(x)_{\chi_{U}} = \chi_{U}(x) = 0 \neq 1 = \chi_{U}(y) = \Gamma(y)_{\chi_{U}}$$

and so $$\Gamma(x) \neq \Gamma(y)$$.

Moreover, $$\Gamma$$ is initial: the topology on $$X$$ is trivially generated by its open sets $$U$$ which can be expressed as

$$U = \{x \in X : \chi_U(x) = 1 \} = \{x : \Gamma(x)_{\chi_U} = 1 \} = \Gamma^{-1}(\prod_{f \neq \chi_U}S \times \{1\})$$

and so $$\Gamma|^{\Gamma(X)} : X \to \Gamma(X)$$ is a homeomorphism with $$\Gamma(X)$$ a subspace of $$S^\mathcal{C}$$.

Thoughts? If correct, I feel like my argument ended up being rather clumsy so I would appreciate any suggestions on how to improve it.

You can use the embedding theorem: if $$X$$ has a family of continuous maps $$f_i: X \to X_i$$ such that the $$f_i$$ separate points and they separate points and closed sets, then $$e= \nabla_i f_i: X \to \prod_{i \in I} X_i$$, defined by $$\pi_i \circ (\nabla_i f_i) = f_i$$ (or $$e(x) = (f_i(x))_{i \in I}$$ more intuitively) is an embedding from $$X$$ into $$\prod_{i \in I} X_i$$. This is proved in Engelking (2.3.20, as the Diagional Theorem) and Willard (8.16) (and it's implicit in an exercise in Munkres somewhere).

Here the $$f_i, i \in I$$ separate points iff for all $$x_1 \neq x_2 \in X$$ there is some $$i \in I$$ such that $$f_i(x_1) \neq f_i(x_2)$$.

The $$f_i, i \in I$$ separate points from closed sets iff for all $$x \in X$$ and all $$C \subseteq X$$ closed with $$x \notin C$$ there exists $$i \in I$$ such that $$f_i(x) \notin \overline{f_i[C]}$$.

Now, if $$S$$ is the Sierpinski space with open $$0$$, we can define for each $$O \subseteq X$$ the characteristic function $$f_O: X \to S$$ defined by $$f_O(x) = 0$$ for $$x \in O$$, $$f_O(x) =1$$ for $$x \notin O$$ and note that $$f_O$$ is continuous, as $$f_O^{-1}[\{0\}] = O$$, so the only non-trivial open has open pre-image.

It is then easy to check that the functions $$f_O:X \to S$$, where $$O$$ ranges over all open subsets of $$X$$ (or if we want to be more minimal, over the $$O$$ in some fixed base $$\mathcal{B}$$ for the topology of $$X$$) has the property that it separates points of $$X$$ if $$X$$ is $$T_0$$ and always separates points from closed sets.

The embedding theorem then ensures that a $$T_0$$ $$X$$ embeds into some product of Sierpinski spaces.

You're argument is trying to reprove parts of the general embedding theorem in a special case, I think.

The theorem is most often used to see that Tychonoff spaces are embeddable into Tychonoff cubes $$[0,1]^I$$, which easily leads to the existence of compactifications for all Tychonoff spaces.

• Great, I didn't know about the embedding theorem (except for its name). Awesome answer as always. Oct 14, 2018 at 8:07