Let $X$ and $Y$ be reasonable spaces. Since $\mathbb{R}^{\infty}$ is contractible, $$ X \times \mathbb{R}^{\infty} \cong Y \times \mathbb{R}^{\infty} \;\;\; \implies \;\;\; X \simeq Y. $$

Is the converse also true?

My vague intuition: the factors of $\mathbb{R}^{\infty}$ provide so much extra room that there will never be a geometric obstruction to producing a homeomorphism. Evidently, there is no homotopy-theoretic obstruction, so maybe the converse is true.

On the other hand, I really have no idea and could be missing something basic. For example, the plane with two punctures is homotopy equivalent to a wedge of two circles. However, I do not know about a homeomorphism $$ (\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}. $$

**Clarification about the meaning of $\mathbb{R}^{\infty}$ and the intent of the question**

When I wrote the question, I had in mind the infinite union $\cup_n \mathbb{R}^n$ inside the product $\mathbb{R}^{\mathbb{N}}$. However, since I would like the answer to the question to be "yes," I am also interested in other versions of $\mathbb{R}^{\infty}$.

I asked the question because of an algebraic limiting construction in a paper I'm writing, and I felt that a topological version of the limit would be satisfying. The algebraic version is already working for spaces like $X \times \mathbb{C}^n$ for large-enough $n$, and converges algebraically to some limiting group, but this doesn't give too many hints about the topology I should use on $\mathbb{C}^{\infty}$, or if the limit can even be considered topologically.

My application involves singular homology, and the direct limit topology is well-suited to this application, but other choices may be as well.

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