Every vector topology on $X\times Y$ is the product of the inherited topologies?

Question

Let $X,Y$ be real (or complex) vector spaces. Let $T$ be a vector topology on $X\times Y$.

The spaces $X$ and $Y$ inherit vector topologies $T_X$ and $T_Y$ from $T$ when we identify $X$ with $X\times\{0\}$ and $Y$ with $\{0\}\times Y$.

Indeed, $T_X = \{G\cap X\,:\, G\in T\}$, where $G\cap X:=G\cap(X\times\{0\}) = \{x\,:\, (x,0)\in G\}$. Analogously for $T_Y$.

They determine the product topology $T_\times$ on $X\times Y$ (whose basis is $\{U\times V\,:\, U\in T_X,\ V\in T_Y\}$).

It is easy to see that the product topology $T_\times$ is finer than $T$. Is it necessarily the same?

(A.) In general? (B.) When we assume (in addition) that $T$ is Hausdorff? (C.) In any major special cases?

Vector topology = + and $\cdot$ are continuous.

[Somebody might write this as: "Is every vector topology the product of vector topologies (on every finite factorization of the space)?"]

[Note: $X,Y,T$ are given, and we ask "(A.)" and "(B.)"; that is, whether necessarily $T=T_\times$.]

[Summary of answers, as of 2025-1-23: This answers (A.,B.) negatively (even in $\ell^2$) and provides several answers to (C.). This answers (A.,B.) negatively (even in $\ell^2$) even without the axiom of choice. Also the third answer seems correct (Banach space counter-example).]

Answer 1

For an explicit counterexample without the axiom of choice, let $(H, \|\cdot\|_H)$ be an infinite-dimensional Hilbert space, and let $E,F$ be two closed subspaces such that $E \cap F = 0$ and $E + F$ is not closed. These can be constructed explicitly; see for instance The direct sum of two closed subspace is closed? (Hilbert space) or Sum of closed spaces is not closed. Let $V = E+F$ which we can also view as $E \times F$ because it is a direct sum. Consider the following two topologies on $V$, which indeed are both Hausdorff vector topologies:

• The topology $T_1$ induced by the norm $\|\cdot\|_H$ restricted to $V$.

• The topology $T_2$ induced by the orthogonal sum norm $|x+y|^2 = \|x\|_H^2 + \|y\|_H^2$, for $x\in E, y \in F$.

Each of these two norms agrees with $\|\cdot\|_H$ when restricted to $E$ or $F$, so $T_1$ and $T_2$ both induce the same subspace topology on $E$ and on $F$. Recall that two topologies induced by norms are equal if and only if the norms are equivalent. But the norms $\|\cdot\|_H$ and $|\cdot|$ are clearly not equivalent, as the latter is complete and the former is not. So $T_1,T_2$ are two different topologies, and they cannot both equal the product topology.

(Actually it is not hard to see that in fact $T_2$ is the product topology, and so $T_1$ is not.)

Answer 2

As I said in a comment, your question is mostly a duplicate. Here are the details.

Lemma. Suppose that $X, Y$ are Hausdorff (actually, $T_1$ is enough) topological spaces and $Z=X\times Y$ (with the product topology). Then for every $p\in X, q\in Y$ the subsets $$ Y_p:=\{p\}\times Y, X_q:=X\times \{q\}\subset Z $$ are closed subsets of $Z$.

Proof. The coordinate projections $\pi_X: Z\to X, \pi_Y: Z\to Y$ are continuous (by the definition of the product topology). Then $$ Y_p=\pi_X^{-1}(p), X_q= \pi_Y^{-1}(q) $$ are both closed in $Z$ since singletons are closed in $X$ and $Y$. qed

Now, suppose that $(V,T)$ is an indecomposable Hausdorff infinite-dimensional TVS, i.e. one which is not a direct sum of two closed infinite-dimensional linear subspaces. (See for instance references given in the answer here for the existence of indecomposable Banach spaces.) Then, by the lemma, $V$ is not (continuously) isomorphic to the sum of two infinite-dimensional topological vector spaces (with the product topology). At the same time, using the Axiom of Choice and, hence, existence of a Hamel basis in $V$, as for every infinite-dimensional vector space, one finds an algebraic decomposition $V=X\oplus Y$ as the direct sum of infinite-dimensional linear subspaces. (Take a Hamel basis $C$ of $V$, represent it as a disjoint union of two infinite subsets $A\sqcup B$ and then take $X, Y$ to be linear spans of $A, B$ in $V$.) Thus, if $T_X, T_Y$ denote the induced topologies on $X$ and $Y$ (as in your question), $(V,T)\ne (X,T_X)\oplus (Y,T_Y)$.

Lastly, if you do not accept the Axiom of Choice, I do not know what to say and I leave it to logicians to deal with this question.

Answer 3

(A.) No. (B.) No. (C.) (1.) and (6.) below. (not even if $X\times Y$ is a separable Hilbert space, whether real or complex) (proof: "(3.)" below).

Some further related results are given below, but it seems to me that not much more is true. All this holds for both real and complex scalars.

(0a.) $T\subset T_\times$, as mentioned in the question.

(0b.) $T=T_\times$ iff the projections $P_X:X\times Y\to X$ and $P_Y:X\times Y\to Y$ are continuous.
Proof: $T_\times$ is the coarsest topology in which $P_X,P_Y$ are continuous, QED.

Rudin: Functional Analysis, Theorem 5.16 (together with (0a.,0b.)) proves (1.) and (2.) below:

(1.) If $X,Y$ are closed and $X\times Y$ is an F-space, then $T=T_\times$.

(2.) If $T=T_\times$ and $T$ is Hausdorff, then $X,Y$ are closed.

(3.) Example (counters (A.,B.)). Set $X=c_c$, $X\times Y=\ell^2$. Then $X$ is not closed (being dense), hence then $T\subsetneq T_\times$, by (2.). Thus, the answers to (A.) and (B.) are no.
(If you extend a basis of $X$ to a basis of $X\times Y$, the span of these extending vectors can be used as the $Y$.)

So the best we can get seems to be the well-known fact that if $X\times Y$ is an F-space, then $T=T_\times$ iff $X$ and $Y$ happen to be closed. It seems that not much more than (1.-3.) is true:

(4.) Example. Let $T=\{\varnothing,X\times Y\}$ (a non-Hausdorff TVS). Then $X,Y$ are not closed but yet $T=T_\times$.

(5.) Example. Let $X=Y={\mathbb R}$ but use the seminorm $(x,y)\mapsto|x+y|$ to define $T$ (a non-Hausdorff TVS). Then $\overline{\{0\}}={\mathbb R}(1,-1)$. Therefore, $T\subsetneq T_\times$.

(6.) However, if $T$ is Hausdorff and $X,Y$ are finite-dimensional, then $T=T_\times$.

Proof: Then $T$ is necessarily the Euclidean topology, because it is the only Hausdorff TVS topology on a finite-dimensional vector space, by Rudin: FA, 1.21. Hence, so are $T_X,T_Y$. Hence, $T=T_\times$, QED.

[(9.) Let me know if somebody wants a proof for $T\subset T_\times$.]