# Continuous functions proof

Let $f$ and $g$ be continuous functions on R with the absolute value metric, and let $S$ $\subset$ R be countable. Show that if $f(x)=g(x)$ for all $x\in S^c$ (complement of $S$), then $f(x)=g(x)$ for all $x\in$ R.

Now since $f$ and $g$ are continuous on R, then they are obviously continuous on $S$ and $S^c$, considering R is the union of both. For $f(x)=g(x)$ to be true for all $x$ in R, then it should be sufficient to show that it holds in $S$, because it is given that it is true for $S^c$. I know I have to use the fact that $S$ is countable, but I can't think of how and why. Any hints?

Thank you.

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Hint: If $f(x)\ne g(x)$, then there is an open interval $O$ containing $x$ so that $f(z)\ne g(z)$ for all $z\in O$. –  David Mitra Dec 9 '12 at 0:01
@DavidMitra So if $S^c$ is dense in R and I assume, by contradiction, that $f(x)$ is not equal to $g(x)$ for all $x$ in R. I would construct the open interval you have stated above. But since $S^c$ is dense, there exists an $x$ in $S^c$ in the open interval. And by hypothesis $f(x)=g(x)$ for all x in $S^c$, I will then arrive at a contradiction. IS this correct? –  Alti Dec 10 '12 at 2:23
Yes, that's correct. You could argue a bit more simply (no need to talk about denseness; though it's ok to do so): $S$ is countable while the open interval is not. It follows that the open interval contains a point of $S^c$. –  David Mitra Dec 10 '12 at 2:55
That was much simpler, thank you! –  Alti Dec 10 '12 at 4:07

For $x\in S$, every $\epsilon$-neighbourhood is uncountable, hence contains points of $S^c$. Thus we can find a sequence $x_n\to x$ with all $x_n\in S^c$. As $f(x_n)-g(x_n)=0$ for all $x_n$ and $f-g$ is continuous, we also have $f(x)-g(x)=0$.

Actually, we don't need that $S$ is countable, but that $S^c$ is dense (which follows from $S$ countable). But if $S$ is uncountable, the claim need not follow (as $S^c$ need not be dense in $\mathbb R$).

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I just to make sure, is $S^c$ dense because uncountable sets are dense? –  Alti Dec 9 '12 at 1:27
@Alti No $\mathbb R_{>0}$ is uncountable but not dense (e.g. there is no point near $-1$) –  Hagen von Eitzen Dec 9 '12 at 9:54
Then why is $S^c$ dense? –  Alti Dec 9 '12 at 20:27
@Alti because $S$ is countable. –  Hagen von Eitzen Dec 9 '12 at 20:46
@Alti For $\epsilon>0$, the $\epsilon$ neighbourhood of $x\in\mathbb R$ is uncountable, hence cannot be a subset of countable $S$, hence intersects $S^c$. –  Hagen von Eitzen Dec 10 '12 at 15:55

Since $S$ is countable, $S^c$ is dense. This means $\overline {S^c} = \mathbb R$. Now proof: if two continuous functions coincide on a dense set, they are equal: For that, we can use, that every metric space is also Hausdorff and given that, we know, that the diagonal $\Delta = \{ (x,x) \in \mathbb R \times \mathbb R \, \mid \, x \in \mathbb R \}$ is closed in $\mathbb R\times \mathbb R$. Consider the function $h\colon \mathbb R \to \mathbb R \times \mathbb R, \: h(x) = \bigl(f(x), g(x)\bigr)$. $h$ is continuous, because $f$ and $g$ are. Thus $h^{-1}(\Delta) = \{ x \in \mathbb R \, \mid \, f(x) = g(x) \}$ is closed. Because of $S^c \subset h^{-1}(\Delta)$ we get $\overline {S^c} = \mathbb R \subset h^{-1}(\Delta)$.

I like this proof because it holds for all Hausdorff spaces. I think i've read it in Munkres Topology book.

Hope this helps, Maikel

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What you want is the following:

DEFINITION A set $D\subseteq \Bbb R$ is dense over $\Bbb R$ if for every $x\in \Bbb R$ there exists a sequence $\{x_n\}\subset D$ such that $x_n\to x$.

DEFINITION Given a function $f:X\to Y$, the restriction of $f$ to $D\subset X$ is the function $f\mid _D:D\to Y$ such that $f\mid_D(x)=f(x)$ for each $x\in D$.

THEOREM Let $f$ and $g$ be continuous over $\Bbb R$. Let $D$ be dense over $\Bbb R$. If $f\mid_D=g\mid_D$, then $f=g$. That is, if $g$ and $f$ agree on every point of $D$, then they agree on every point of $\Bbb R$.

PROOF Let $x\in \Bbb R\setminus D$. Then there exists a sequence $\{x_n\}\subset D$ such that $x_n\to x$. And for this sequence $\lim\; f(x_n)=f(x)=g(x)=\lim\; g(x_n)$, so the claim follows. Note that since $\{x_n\}\subset D$, $x_n\neq x$ for each $n$.

The proof is rather simple if you have already characterized continuity in terms of sequences, that is

PROPOSITION Let $f:[a,b]\to \Bbb R$ be continuous. Then $f$ is continuous if, and only if, for each sequence $\{x_n\}\subset [a,b]$ such that $x_n\to x$ (with $x\neq x_n$ for each $n$) we have that $f(x_n)\to f(x)$.

PROOF Suppose $f$ is continuous, and $x_n\to x$. Then for each $\epsilon>0$ we get a $\delta >0$ such that for each $y$, whenever $|x-y|<\delta$, $|f(x)-f(y)|<\epsilon$. Since $x_n\to x$, we can obtain an $N$ such that if $n>N$, then $|x-x_n|<\delta$, so that for $n>N$; $|f(x)-f(x_n)|<\epsilon$, and one direction follows. Can you prove the other direction?

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