Let $X$ be the product of the continuum discrete two-point spaces. Show that not every sequence in $X$ has a convergent subsequence.
We can show that $X$ is compact by the Tikhonov theorem. And I suppose $X$ must be homeomorphic to the Cantor set.
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2Are "colons" discrete two-point spaces? I'd not heard "colon" as name for a topological space... – paul garrett Feb 07 '18 at 23:29
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No, the Cantor set is homeomorphic to a product of countably many copies of ${ 0, 1 }$. In particular, $X$ has the wrong cardinality $2^c$ to be homeomorphic to the Cantor set which has cardinality $c$. – Daniel Schepler Feb 07 '18 at 23:33
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Yes, you're right about colons. I am sorry. I fixed. – SWarn Feb 07 '18 at 23:35
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I explained the same idea as @DanielSchepler in this answer – Henno Brandsma Feb 08 '18 at 21:51
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$X$ is called a "Cantor cube", as are all spaces of the form ${0,1}^\kappa$ where $\kappa$ is a cardinal number. Only for $\kappa$ countable do we get a homeomorphic space to the standard Cantor set in $[0,1]$. – Henno Brandsma Feb 08 '18 at 21:54
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Hint: you should presumably know that in a product space, a sequence $(x_n)_{n=1}^\infty$ converges to $x$ if and only if $x_n(i) \to x(i)$ pointwise for each $i$ in the index set of the family. So, the goal is to produce a sequence such that after taking any subsequence, you still have oscillation of values of $x_n$ at some $i$.
So, to come up with a concrete example, let's identify the index set with $P(\mathbb{N})$, the set of subsets of $\mathbb{N}$. Then consider the sequence $(x_n)_{n=1}^\infty$ of elements of $P(\mathbb{N}) \to \{ 0, 1 \}$ defined by: $$x_n(S) = \begin{cases} 1, & n \in S \\ 0, & n \notin S. \end{cases}$$ Now, suppose you have a subsequence $(x_{n_k})_{k=1}^\infty$ of this sequence. Your goal from here will be to find an index $S \in P(\mathbb{N})$ such that $x_{n_k}(S)$ is not eventually constant, and therefore the subsequence cannot have a limit because $x_{n_k}(S)$ does not have a limit.
Daniel Schepler
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Pictorially, if you instead use the interval $[0,1]$ as the index set, then the sequence roughly corresponds to the indicator functions of $[0, 1/2]$, $[0, 1/4] \cup [1/2, 3/4]$, $[0, 1/8] \cup [1/4, 3/8] \cup [1/2, 5/8] \cup [3/4, 7/8]$, etc. – Daniel Schepler Feb 08 '18 at 00:01
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I am sorry, but what do we call the limit of the sequence from the zeroes and the units? – SWarn Feb 08 '18 at 19:29
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A sequence $(x_n)_{n=1}^\infty \subseteq { 0, 1 }$ converges to $L \in { 0, 1 }$ if and only if $x_n = L$ for sufficiently large $n$. (Pretty much the same statement is true for convergence of sequences in any discrete space.) Is that what you were asking? – Daniel Schepler Feb 08 '18 at 19:45
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Yes, thank you. Maybe I'm dull, but from compactness follows sequential compactness. So every sequence has a convergent subsequence. – SWarn Feb 08 '18 at 20:05
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Compactness only implies sequential compactness for metric spaces, while ${ 0, 1 }^c$ is not metrizable. – Daniel Schepler Feb 08 '18 at 20:09
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@Shorty12319 No, in general spaces compactness does not imply sequential compactness. ${0,1}^\mathfrak{c}$ is one of the counterexamples, as is $\beta \omega$. – Henno Brandsma Feb 08 '18 at 21:49
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@DanielSchepler Even for $X$ sequential (much weaker than metric) we get that compactness implies sequential compactness. – Henno Brandsma Feb 08 '18 at 21:56
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@ Henno Brandsma, can you explain please why $ { 0, 1 }^c$ is one of the counterexamples? I thought it is metrizable. – SWarn Feb 08 '18 at 22:16