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$X$ is a topological space of infinite cardinality which is homeomorphic to to $X \times X$.Then

a) $X$ is not connected

b) $X$ is not compact

c) $X$ is not homeomorphic to a subset of $\mathbb{R}$

d) None of the above


the answer is d and i don't know how $\mathbb{N}$ is a counterexample of a,b and c. Now under discrete topology it is homeomorphic to $\mathbb{N}$ since any bijective function from $\mathbb{N}$ to $\mathbb{N}\times \mathbb{N}$ is continuous and inverse also continuous under discrete topology (inverse map of any open subset is open). But how consider a and b???

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  • $\begingroup$ A discrete space that is also compact must be finite. For connectedness - look for a disconnection - representation of the entire space as a disjoint union of open sets. $\endgroup$ Sep 1 '18 at 10:15
  • $\begingroup$ yea it is compact also but a space with the discrete topology is connected only if it is empty or contains exactly one point.so can we show it connected in other any topology ?? $\endgroup$
    – RAM_3R
    Sep 1 '18 at 10:51
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As you said, $\mathbb{N}$ is a counterexample for c).

Let $Q = \prod_{n=1}^\infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q \times Q \approx Q$.

Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $\mathbb{R}$, that is, homeomorphic to an interval $J$. But $J \times J$ is not homeomorphic to $J$.

$Q$ is a counterexample for both a) and b), and the Cantor set (see Henno Brandsma's answer) is a counterexample for both b) and c).

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  • $\begingroup$ thanks nice example . But can $\mathbb{N}$ be connected and compct in any other topology ??? $\endgroup$
    – RAM_3R
    Sep 1 '18 at 10:46
  • $\begingroup$ No. It is known that any compact connected space must be uncountable. $\endgroup$
    – Paul Frost
    Sep 1 '18 at 10:56
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    $\begingroup$ I think you mean that a (non-empty) compact connected Hausdorff space is uncountable. More generally, a non-empty locally compact Hausdorff space with no isolated points has cardinal at least $2^{\aleph_0}.$ $\endgroup$ Sep 1 '18 at 21:11
  • $\begingroup$ @DanielWainfleet You are right! It seems that there are two "schools" with a different interpretation of the word "compact". One school (I guess mainly popular in Europe) defines a space $X$ to be quasicompact if every open cover has a finite subcover. If $X$ is in addition Hausdorff, it is called compact. This is my understanding. The other school uses the wording "compact" instead of "quasicompact" and "compact Hausdorff" instead of "compact". In textbooks you will find both variants, the first for example in Dugundji and Engelking. $\endgroup$
    – Paul Frost
    Sep 1 '18 at 22:51
  • $\begingroup$ @DanielWainfleet By the way, each connected Tychonoff space (= completely regular $T_1$-space) containing at least two points has cardinality at least $2^{\aleph_0}$. See Engelking 6.1.3. $\endgroup$
    – Paul Frost
    Sep 1 '18 at 22:56
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It's d);

a) is refuted by $X = \mathbb{R}^\omega$. b) and c) by the Cantor set.

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  • $\begingroup$ yea thank you but i want to know about$\mathbb{N}$ in other topological space like indiscrete topology ; is it connected??? $\endgroup$
    – RAM_3R
    Sep 1 '18 at 12:05
  • $\begingroup$ Another example for a) is a separable Hilbert space. $\endgroup$ Sep 1 '18 at 12:26
  • $\begingroup$ @KaviRamaMurthy That's homeomorphic to $\mathbb{R}^\omega$. $\endgroup$ Sep 1 '18 at 12:27
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    $\begingroup$ @RAM_3R $X=\mathbb{N}$ in the indiscrete topology is compact, connected and obeys $X\times X \simeq X$, is that your question? It’s not a subspace of the reals of course. $\endgroup$ Sep 1 '18 at 14:58

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