The converse of Tychonoff theorem?

Suppose, a topological space $(X, \mathscr{T})$ is compact, and the cardinality of $X$ is $2^\kappa$. Is there a compact space $A$ with the cardinality not larger than $2^\kappa$, such that $A^\kappa$ with the product topology $\mathscr{T'}$coincide with $(X, \mathscr{T})$? In other words, Could $(X, \mathscr{T})$ be homeomorphic to $(A^\kappa, \mathscr{T'})$?

Added: As pointed out in Ittay Weiss's and Nate Eldredge's answers, the answer is negative when $\kappa$ is finite and when $(X, \mathscr{T})$ contains a isolated point. I become interested in the condition that guarentees the homeomorphism exists. In particular, I'm interested in the case when $\kappa$ is infinite, and no isolated point in $(X, \mathscr{T})$, say, the closed interval with the usual topology.

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If $A$ contains at least two points and $\kappa$ is infinite, then it is easy to see that every open set in $A^\kappa$ is infinite; in particular $A^\kappa$ has no isolated points. But of course we can construct a compact space of cardinality $2^\kappa$ that does have isolated points; for instance the disjoint union of $\{0,1\}^\kappa$ (with its product topology) and one additional point. Therefore such a space cannot be homeomorphic to $A^\kappa$ for any $A$.

Regarding the case $X = [0,1]$: it is not homeomorphic to any product $A^\kappa$. Suppose it were. Since $A$ is the continuous image of $X$ under the coordinate maps, $A$ is compact and connected. $A$ is also homeomorphic to a subset of $X$ under any map like $x \mapsto (x, x_0, x_0, \dots)$. But the only compact connected subsets of $[0,1]$ are: the empty set, singletons, and closed intervals. The first two cases are absurd, and it's easy to see that $[0,1]$ is not homeomorphic to $[0,1]^\kappa$ for any $\kappa > 1$, since $[0,1]^\kappa$ remains connected when any one point is deleted.

I suppose one might ask for necessary and sufficient conditions for a compact space $X$ to be homeomorphic to a product. I don't know what they might be.

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Thank you for your answer. What can we say about the compact space with no isolated point? –  Metta World Peace Jan 20 '13 at 5:01
A relevant theorem: if $X$ is first countable and a zero-dimensional Tychonoff space, then $X^\omega$ (and thus all higher powers) is (are) homogeneous. So this quite limits the compact spaces homeomorphic to some infinite product of first countable spaces: they must be homogeneous (so this also handles $[0,1]$ e.g.) –  Henno Brandsma Jan 20 '13 at 8:11

The answer is no. Very simple counter examples can be constructed as follows. Consider the cardinal $4=2^2$. Every finite topological space is compact. There are four topologies on a set with two elements. That means that there are at most $4$ spaces of cardinality $4$ that are of the form $A^2$ for some $2$-point space $A$. However, there are 355 distinct topologies on a four point set. So, certainly many many $4$-set topologies are not of the form $A^2$ for any $2$-set topology $A$. (Note that relaxing the question here to $X$ being a product up to homeomorphism or even homotopy is still answer negatively by a similar counting argument.)

For infinite counterexamples, here is an algebraic topology argument. It is well known that for any finitely generated group $G$ there exists a compact CW-space $X$ whose fundamental group is $G$. The fundamental group functor respects arbitrary products: $\pi_1(\prod Xi)\cong \prod\pi_1(X_i)$. So, now just take $G$ to be any non-trivial simple group. If the corresponding $X$ would be a product of $\kappa$ many spaces than $G$ would be a non-trivial product and thus not simple.

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Thank you for your answer. Unfortunately, I know nothing about algebraic topology. –  Metta World Peace Jan 20 '13 at 5:05

Assume that $X$ is infinite. If $X$ is homeomorphic to $Y^\kappa$ for some infinite $\kappa$, then $X\cong Y^\omega\times Y^\kappa$, and $Y^\omega$ has cardinality at least $2^\omega$. For each $y\in Y^\omega$ let $X_y=\{y\}\times Y^\kappa$; then $X_y\cong X$, so $\{X_y:y\in Y^\omega\}$ is a partition of $X$ into $|Y|^\omega$ closed subsets homeomorphic to $X$.

Now let $\kappa$ be any infinite cardinal. Yet $Y=\{0,1\}^\kappa$ with the product topology, where $\{0,1\}$ is discrete, let $D_\kappa$ be a discrete space of cardinality $2^\kappa$, let $X=D_\kappa\times Y$, and let $X^*$ be the one-point compactification of $X$. Then $X^*$ is a compact Hausdorff space of cardinality $2^\kappa$ without isolated points, and the point at infinity is the unique point of $X^*$ whose character (i.e., minimum cardinality of a local base) is $2^\kappa$, so $X^*$ does not contain even two subspaces homeomorphic to itself. In particular, $X$ is not homeomorphic to $Z^\kappa$ for any space $Z$.

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Thank you for your answer.But I have troube in understanding it.$D_\kappa$ and $Y$ are compact, implies that their product $X$ is compact. So $\{\infty\}$ is open in $X^{*}$, since its complement $X$ is compact, which means $\infty$ is a isolated point. What's wrong? –  Metta World Peace Feb 3 '13 at 5:12
@MettaWorldPeace: No, $D_\kappa$ is an infinite discrete space, so it’s very far from being compact. –  Brian M. Scott Feb 3 '13 at 5:13
I see. It's indeed an ingenious construction, though, to me,it's like something comes out of nowhere. Thank you very much. –  Metta World Peace Feb 3 '13 at 5:26
@MettaWorldPeace: You’re welcome. (It’s funny that you should say that: to me it seems very natural to try to make one point different from all the rest!) –  Brian M. Scott Feb 3 '13 at 5:27