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.  
 A: 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 $\mathbb R$).
A: 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
A: 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? 
