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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Prove that if $S$ is a dense (i.e., closure of $S$ is $\mathbb{R}$) subset of $\mathbb{R}$ such that $f(s) = 0$ for every $s$ in $S$, then $f(x) = 0$ for every $x$ in $\mathbb{R}$.

I think the following may be helpful:

Let $f$ be a function with domain $E$ and fix $P\in E$. The function $f$ is continuous at $P$ if and only if for every sequence $\{a_{j}\}\subseteq E$ satisfying $\lim_{j \to \infty} a_{j} = P$ it holds that $\lim_{j \to \infty} f(a_{j}) = f(P)$

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    the set $\{x|f(x)\neq 0\}$ is open, $S$ is dense2012-02-23
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    A great argument! (@Blah) Alex, your idea is probably more typical of such situations and one can continue that way, but I just love Blah's argument. There are no sequences here, no metric, just pure topology and all in one sentence.2012-02-23
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    I am almost certain this is a duplicate, but I cannot find the question it is a duplicate of.2012-02-23
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    You can try to prove the more general thing: $f(\overline{A}) \subset \overline{f(A)}$ for continuous $f$ and every $A \subset \mathbb{R}$ (it can even be taken as a definition of continuity).2012-02-23
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    @savick01 I agree that Blah's argument is nicer, but I didn't want to use general topological arguments because I suspect that the OP is more comfortable with sequences in $\mathbb R$, and wanted to use the OP's idea.2012-02-23
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    @AlexBecker: I'm not attacking your answer, I think it was posted after my comment anyway. I think the best we could've done was to give both of the answers. If the OP is not familiar with topology, then we should give them an understandable solution, especially if they've made the first step towards them. But it is always good to show (and to learn!) the simplest solution to a particular problem (in that case Jacob's answer which actually is Blah's comment uses no topological word not included in the question) using the tools of the area of mathematics mentioned in the tag list.2012-02-23
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    @savick01 Ah, sorry to misread your comment. Blast this common name!2012-02-23

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