It is known that no two distinct finite powers of the closed unit interval are homeomorphic:

$I^m$ is homeomorphic to $I^n$ iff $m=n$. (Brouwer, Lebesgue, 1911)

Is the analogous result for infinite powers of $I$ true?

That is, is it true that $I^\alpha$ is homeomorphic to $I^\beta$ iff $\alpha=\beta$, for cardinal numbers $\alpha$ and $\beta$?

(An affirmative answer would give a hope of defining a transfinite inductive dimension for all spaces, a problem essentially posed by Carl Menger.)

If not, what are the homeomorphism classes among the infinite powers of $I$?

Added Later: Since $|I^\alpha|=2^{\aleph _0 \alpha}=2^\alpha$ whenever $\alpha$ is infinite, assuming the Generalised Continuum Hypothesis, of course, $2^\alpha=2^\beta \Rightarrow \alpha=\beta$ and $I^\alpha$ and $I^\beta$ have different cardinalities if $\alpha$ and $\beta$ are distinct. But the answer below shows that they can be proved to be not homeomorphic within ZFC, and, more interestingly, even when they have the same cardinality (in some model which violates GCH).


1 Answer 1


Yes. $I^{\alpha}=I^{\beta}$ does imply that $\alpha=\beta$. To see this, suppose that $\alpha$ is an infinite cardinal. Then each point in $I^{\alpha}$ is the intersection of $\alpha$ many open sets, but each point in $I^{\alpha}$ is not the intersection of less than $\alpha$ many open sets.

In greater detail, if $(x_{\beta})_{\beta<\alpha}\in I^{\alpha}$, then for each $n\in\mathbb{N}$ and $J\subseteq\alpha$ with $|J|<\aleph_{0}$, let $U_{J,n}=\{(y_{\beta})_{\beta<\alpha}:|y_{j}-x_{j}|<\frac{1}{n}\,\textrm{for}\,j\in J\}$. Then $(x_{\beta})_{\beta<\alpha}=\bigcap_{J,n}U_{J,n}$, so $(x_{\beta})_{\beta<\alpha}$ is the intersection of $\alpha$ many open sets.

Now, suppose $|J|<\alpha$ and $U_{i}\subseteq I^{\alpha}$ is an open neighborhood of $(x_{\beta})_{\beta<\alpha}$ for $i\in J$. Then for each $i\in J$ there is a finite $J_{i}\subseteq\alpha$ where if $(y_{\beta})_{\beta\in\alpha}\in I^{\alpha}$ and $y_{j}=x_{j}$ for $j\in J_{i}$, then $(y_{\beta})_{\beta\in\alpha}\in U_{i}$. Since $|\bigcup_{i\in J}J_{i}|=|J|<\alpha$, $\bigcup_{i\in J}J_{i}\neq\alpha$. Therefore, whenever $y_{\beta}=x_{\beta}$ for $\beta\in\bigcup_{i\in J}J_{i}$, we have $(y_{\beta})_{\beta\in \alpha}\in\bigcap_{i\in J}U_{i}$, but we do not necessarily have $(y_{\beta})_{\beta\in \alpha}=(x_{\beta})_{\beta<\alpha}$. Hence, $\bigcap_{i\in J}U_{i}$ is not equal to $\{(x_{\beta})_{\beta<\alpha}\}$.

  • $\begingroup$ Great answer and fast enough too, though I was a bit confused by the use of $I$ in the last paragraph (and have suggested editing it to $J$). This answer also makes it clear exactly which cubes are perfectly normal etc., among other things, in an elegant way (I have not seen a direct proof in textbooks). Is there a reference? $\endgroup$ Jun 3, 2014 at 17:40
  • $\begingroup$ I do not have a direct reference, but I simply found that $\alpha$ can be extracted from $I^{\alpha}$ by a cardinal invariant which for compact spaces is equivalent to the local character of a space. One should therefore look up cardinal invariants of a topological space for similar invariants that can be used to distinguish topological spaces. $\endgroup$ Jun 3, 2014 at 18:52
  • $\begingroup$ Can you specify that cardinal invariant? (I could not find one.) $\endgroup$ Jun 3, 2014 at 22:15
  • 8
    $\begingroup$ For an infinite cardinal $\alpha$, the cube $I^{\alpha}$ has many cardinal invariants equal to $\alpha$: weight, character, pseudo-character (as in Joseph´s answer), spread, tightness, and many more. You can look at Hodel´s article in the Handbook of Set Theoretic Topology for more information on those. $\endgroup$ Jun 5, 2014 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.