Every separable metric space has cardinality less than or equal to the cardinality of the continuum

Asked Jul 21 '22 at 12:12

Active Jul 21 '22 at 12:19

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I want to prove "Every separable metric space has cardinality less than or equal to the cardinality of the continuum".

$\textbf{I want to approach by sequences.}$

My attempt: Let $S$ be a countable dense subset of a metric space $X$. By density property for each $x\in X$ there is a sequence $\{x_n\}$ in $S$ such that $x_n \rightarrow x$.

I want to define a injection from $X$ into $\mathbb{R}^S$. How can I?

If I can then we get easily that cardinality of $X$ is atmost that of continuum.

Please help.

edited Jul 21 '22 at 12:19

asked Jul 21 '22 at 12:12

user1234

$x \to (y\to d(x,y))$ – geetha290krm Jul 21 '22 at 12:15
@geetha290krm This includes metric. Can I do it only using sequences? – user1234 Jul 21 '22 at 12:17
@SuzuHirose I am looking for sequence approach. Your link uses metric approach. – user1234 Jul 21 '22 at 12:18
@geetha290krm means, you are considering a map from $S$ into $\mathbb{R}$ via $y \mapsto d(x, y) $ where $x$ is fixed? – user1234 Jul 21 '22 at 15:21
@user1234 See https://math.stackexchange.com/questions/2995939/a-first-countable-separable-hausdorff-space-has-at-most-the-continuum-cardinali – Mason Jul 21 '22 at 20:27

Every separable metric space has cardinality less than or equal to the cardinality of the continuum

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