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When $X$ is an arbitrary topological space, I need to know which of the followings are true or false

$1$. If $X$ is compact, then every sequence in $X$ has a convergent subsequence.

$2$. If every sequence in $X$ has a convergent subsequence then $X$ is compact.

$3$. $X$ is compact iff every sequence in $X$ has a convergent subsequence.

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If you search for sequentially compact here at MSE, you'll find a few related questions. Wikipedia article Sequentially compact space has some useful information too. – Martin Sleziak Oct 25 '12 at 8:41
up vote 5 down vote accepted

All of them are false.

(1) $[0,1]^{[0,1]}$ is compact, but not sequentially compact: On one hand, we have that $[0,1]$ is compact, so its power $[0,1]^{[0,1]}$ is compact by Tychonov. But consider the following sequence \[ x_n = \sum_{k=0}^{2^{n-1}-1} \chi_{[k2^{1-n}, (2k+1)2^{-n})}\colon [0,1] \to [0,1] \] where $\chi_A$ denotes the characteristic function of a set $A$. Let $(x_{n_k})$ an arbitrary subsequence of $(x_n)$, set $t := \sum_{k=0}^\infty 2^{-n_{2k+1}}$. Then $x_{n_k}(t) = 0$ if $k$ is even, but $x_{n_k}(t) = 1$ if $k$ is odd. Hence $(x_{n_k})$ doesn't converge.

(2) $[0, \omega_1)$ fails to be compact, as $[0,\alpha)$, $\alpha < \omega_1$ is an open cover without finite subcover, but each sequence is bounded are has a convergent subsequence.

(3) see (1) and (2).

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It be noted that the statements do hold for metric spaces. – Lord_Farin Oct 25 '12 at 8:40
could you please explain $(1)$ a little? – Un Chien Andalou Oct 25 '12 at 8:41
@Flute: Will do. – martini Oct 25 '12 at 9:25

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