Every separable metric space has cardinality less than or equal to the cardinality of the continuum.
This was stated in my book in a series of questions using the post office metric on $\Bbb R^2$, but I can't think of a way to prove this.
Every separable metric space has cardinality less than or equal to the cardinality of the continuum.
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2If $S$ is a countable dense subset of the space, then every element of the space is a limit of a sequence from $S$. – David Mitra Jun 29 '16 at 13:12
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1What @DavidMitra said, plus $\aleph_0^{\aleph_0} = \mathfrak{c}$. – Henno Brandsma Jun 29 '16 at 15:32
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Could you explain how that shows it can't be greater than the continuum cardinality? I'm contused as to how that relates. – Oliver G Jun 30 '16 at 02:09
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Map $x$ to a sequence from $S$ that converges to $x$. This is injective, so the cardinality of your space is at most the cardinality of the set of sequences from $S$. But this is the same as the cardinality of the set of sequences of positive integers. – David Mitra Jun 30 '16 at 20:46
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Why is the map $x\mapsto (x_n)$ well-defined? In other words, for a fixed $x,$ why can't we have two sequences converging to $x$? – Idonknow Oct 16 '17 at 01:06
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We are choosing such a sequence for each x, we don't claim that it is unique. – moteutsch Apr 12 '18 at 21:01