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


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


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.

  • $\begingroup$ 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 $X_j \subset \mathbb N$). $\endgroup$
    – Paul Frost
    Feb 24 '20 at 23:40
  • $\begingroup$ @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. $\endgroup$ Feb 24 '20 at 23:46
  • $\begingroup$ 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 ;-) $\endgroup$
    – Paul Frost
    Feb 24 '20 at 23:49
  • $\begingroup$ @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. $\endgroup$ Feb 24 '20 at 23:52
  • $\begingroup$ Nice idea. Then we can also omit the assumption that all $X_j$ are non-emoty. $\endgroup$
    – Paul Frost
    Feb 24 '20 at 23:56

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