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Need to know how to prove that a sequence $\{x_k\}_{k=1}^\infty\subset \mathbb{R}^n$ converges to $x$ if and only if the map $ f:\{1,2,3...\} \to\mathbb{R}^n$, $ f(j) = x_j$, is continuous.

It's been puzzling me for the last few hours now. I know that a sequence converges to a point just when every open set containing x contains all but finitely many of the points in the sequence.

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    Please check that I didn't change the meaning.2012-11-13
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    The set $\{1,2,3...\}$ is discrete, so any map $f:\{1,2,3...\} \to\mathbb{R}^n$ is continuous.2012-11-13
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    This is an odd question and I don't think it's true: if you take the naturals $\,\{1,2,3,...\}\,$ with the inherited topology from the reals, then this is a *discrete* space, but *any* function from a discrete space to *any other space* is always continuous...2012-11-13
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    So there's no actual proof? It's enough to say that because the set is discrete then it's obviously going to be continuous?2012-11-13
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    No, that's not a proof, because for instance the function $f:\Bbb{N}\to\Bbb{R}$ given by $f(n)=n$ is continuous but corresponds to the sequence $\{n\}$, which doesn't converge. The claim is false, although it might work with the cofinite topology on $\Bbb{N}$.2012-11-13
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    It won't work with to cofinite topology either, I think. The statement "$x_k \to x$" depends on $x$, but "$f \colon j \mapsto x_j$ is continuous" doesn't.2012-11-13

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