# How to prove that $\Pi_{j\in J}X_j$ is a $T_0$ spaces if and only if each factor $X_j$ is a $T_0$ space

Following a reference from "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias.

Definition

A topological space $$(X,\mathcal{T})$$ is a $$T_0$$ space if for any $$x,y\in X$$ there exist a open set $$U$$ such that $$|U\cap\{x,y\}|=1$$, that is $$U\cap\{x,y\}=\{x\}\lor U\cap\{x,y\}=\{y\}$$.

Well with this definition we prove the following theorem

Theorem

Let be $$\mathfrak{X}=\{(X_j,\mathcal{T}_j):j\in J\}$$ a collection of topological not empty spaces, so the product space $$\Pi_{j\in J}X_j$$ of the collection is $$T_0$$ if and only if any term $$X_j$$ of the product is $$T_0$$.

proof. For starters we suppose that $$\Pi_{j\in j}X_j$$ is a $$T_0$$ space. Well using the Choice Axiom for any $$i\in J$$ we can define for some fixed $$z\in\Pi_{j\in J}$$ the set $$Z_i=\{x\in\Pi_{j\in J}: x(j)=z(j), j\neq i \land x(j)=x_h\in X_i, j=i\}_{h\in|X_i|}$$ and we prove that it is homeomorphic to $$X_i$$. So we consider the restriction $$\pi_i|_{Z_i}$$ of the projection $$\pi_i$$ and we observe that by a previous theorem it is continuous on the subspace topology $$\mathcal{T}_Z$$ of $$Z$$; moreover since two elements $$x$$ and $$y$$ of $$Z_i$$ differ only for their values $$x(i)$$ and $$y(i)$$ it result that $$\pi_i|_{Z_i}$$ is bijective and so it is that $$\forall A\in\mathcal{T}:\pi_i|_{Z_i}(A\cap Z_i)=\pi_i|_{Z_i}(A)\cap\pi_i|_{Z_i}(Z_i)=\pi_i(A)\cap X_i=\pi_i(A)\in\mathcal{T}_i$$ from which we can colude that $$\pi_i|_{Z_i}$$ is open and so it is a homeomorphism between $$Y_i$$ and $$X_i$$: so since any subspace of a $$T_0$$ space is $$T_0$$ space and since the omeomorphism preserve the $$T_0$$ property we can conclude that $$X_i$$ is a $$T_0$$ space for any $$i\in J$$.

Now we suppose that for each $$j\in J$$ it result that $$X_j$$ is a $$T_0$$ space. So if $$x,y\in\Pi_{j\in J}X_j:x\neq y$$ it result that $$I=\{i\in J: \pi_i(x)\neq\pi_i(y)\}\neq\varnothing\Rightarrow(\forall i\in I)\exists A\in\mathcal{T_i}:A\cap\{\pi_i(x),\pi_i(y)\}=\{\pi_i(x)\}\lor A\cap\{\pi_i(x),\pi_i(y)\}=\{\pi_i(y)\}\Rightarrow(\forall i\in I)\exists A\in\mathcal{T_i}:\pi^{-1}_i(A)\cap\{x,y\}=\{x\}\lor\pi^{-1}_i(A)\cap\{x,y\}=\{y\}$$ since differentely it would result or that $$\pi^{-1}_i(A)\cap\{x,y\}=\varnothing\Rightarrow x\notin\pi^{-1}_i(A)\land y\notin\pi^{-1}_i(A)\Rightarrow\pi_i(x)\notin A \land\pi_i(y)\notin A\Rightarrow A\cap\{\pi_i(x),\pi_i(y)\}=\varnothing$$ or that $$\pi^{-1}_i(A)\cap\{x,y\}=\{x,y\}\Rightarrow x\in\pi^{-1}_i(A)\land y\in\pi^{-1}_i(A)\Rightarrow A\cap\{\pi_i(x),\pi_i(y)\}=\{\pi_i(x),\pi_i(y)\}$$ and so by the continuity of the projection $$\pi_i$$ we conclude that $$\Pi_{j\in J}X_j$$ is a $$T_0$$ space.

Well I ask if my poof is correct: in particular I doubt that the demonstration of the "openness" of $$\pi_i|_{Z_i}$$ is uncorrect, since it would be $$\pi_i|_{Z_i}(A\cap Z_i)\neq\pi_i(A)\cap\pi_i|_{Z_i}(Z_i)$$. If the proof is uncorrect, how prove the assertion? So could someone help me, please?

You can indeed use that each $$X_j$$ embeds as a subspace into $$X=\prod_{j \in J} X_j$$, and if you pick a point $$z \in X$$ (using AC, but otherwise $$X$$ is empty and the implication "$$X$$ is $$T_0$$" implies "each $$X_j$$ is $$T_0$$" is false, so AC must be assumed anyway for your theorem to hold) and define for a fixed but arbitrary $$j_0 \in J$$, the map $$e: X_j \to X$$ by $$\pi_{j_0}(e(x))=x$$ and $$\pi_j(e(x))=z_j$$ for $$j \neq j_0$$. Then $$e$$ is continuous by the universal mapping theorem for products: its compositions with projections are either the identity on $$X_{j_0}$$ or constant maps, both of which are continuous always. And $$e$$ is 1-1 and has a continuous inverse $$\pi_{j_0}\restriction_{e[X_j]}$$ so that $$X_j \simeq e[X_j] \subseteq X$$ and so if $$X$$ is $$T_0$$, so is $$X_{j_0}$$, for each index $$j_0$$.

The embedding fact is just a separate fact (using AC) that could be used as a general lemma (nothing to do with $$T_0$$ or any property): each space embeds into a product containing it. Prove it once, use it everywhere.. We cannot use an open projection argument because $$T_0$$ need not be preserved by open continuous maps, or just continuous maps. I don't need to define $$Z_i$$ as you do, considering $$e[X_j]$$ is enough (it's the same thing).

Conversely, if all $$X_i$$ are $$T_0$$ and $$x \neq y$$ are two points of $$X$$, it must be the case there exists at least coordinate $$j_1 \in J$$ such that $$x_{j_1} \neq y_{j_1}$$. In $$X_{j_1}$$ we pick an open set $$O$$ such that $$O$$ contains exactly one of $$x_{j_1}$$ and $$y_{j_1}$$. Then $$O':=\pi_{j_1}^{-1}[O]$$ is open in $$X$$ and if $$O$$ contained $$x_{j_1}$$, $$O'$$ contains $$x$$ and vice versa. Likewise for $$y_{j_1}$$. So $$O'$$ is as required for $$x$$ and $$y$$ (contains exactly one of them), and $$X$$ is $$T_0$$.

Your proof is just "formulae", use more words, I'd say. It's clearer.

• Without AC we cannot say anything about $X$. It may be empty or non-empty. In some special cases we do not need AC to conclude that $X$ is non-empty (for example, if all $X_j \subset \mathbb N$). Feb 24 '20 at 23:40
• @PaulFrost If AC fails we have a collection of non-empty sets with empty product and if we give each of them the indiscrete topology we have a product that is trivially $T_0$ (being empty) and at least one of the factors not $T_0$ so not AC implies a counterexample exists, so we need AC in that sense. Not for every instance of the proof. Feb 24 '20 at 23:46
• You are right, if AC fails, then there exists a collection of non-empty sets with empty product. But not assuming AC does not mean that AC fails. Okay, perhaps a bit nitpicky ;-) Feb 24 '20 at 23:49
• @PaulFrost The embedding fact can better be formulated avoiding the issue: if $X \neq \emptyset$, then $X_j$ embeds into $X$, for each $j$. Make the assumption explicit. Feb 24 '20 at 23:52
• Nice idea. Then we can also omit the assumption that all $X_j$ are non-emoty. Feb 24 '20 at 23:56