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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](http://math.stackexchange.com/search?q=sequentially+compact) here at MSE, you'll fin$d$ a few related questions. Wiki$p$edia article [Sequentially compact space](http://en.wikipedia.org/wiki/Sequentially_compact_space) has some useful information too.2012-10-25

1 Answers 1

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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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    @Flute: Will do.2012-10-25