# Analytic continuation principle and closed sets

Let $f$ and $g$ two analytic function. The two funtion are not equal in the whole complex plane. Due to the analytic continuation principle, any equation of the form $f(z)=g(z)$ cannot holds in any set with an accumulation point, otherwise these functions are equal for all values. Since every point in an open set is an accumulation point, then this equality must hold in a closed set. I have

$f(z)=g(z)$ in the set: $0<Re(z)<1$ and $Im(z)∈ℝ$.

Can I deduce that this hapen only on one line $Re(z)=a$ with $0<a<1$ or there are several lines in which this equality holds.

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The set where they two analytic functions are equal is closed simply because $f,g$ are continuous. – Andrés E. Caicedo Dec 29 '12 at 17:33
Yes, I know that. Can you see my question. – ZE1 Dec 29 '12 at 17:36
This is not very clear, but I think it's due to legitimate confusion that can be cleared up in the answers. I vote not to close. – Brett Frankel Dec 29 '12 at 17:44

I think you are a little confused about accumulation points: If $f(z)$ and $g(z)$ are both defined on some connected open set, and the set where $f=g$ has an accumulation point, then $f$ and $g$ are equal. So if $f$ and $g$ are both defined on all of $\mathbb{C}$ in your example above, then they are indeed equal everywhere.
Likewise, if $f$ and $g$ are both defined on a neighborhood of some line $\ell$ which is not vertical, then $f$ and $g$ agree on $\ell$ because $\ell$ is simply connected and $f=g$ on $\ell\cap \{z: 0>\Re(z)<1\}$. So yes, the lines you listed are the only lines on which you can say for sure that $f=g$ because those are the only lines where you know $f$ and $g$ are defined.
$f$ and $g$ need not agree everywhere, or even on every vertical line, if they are not defined everywhere. Take $f(z)=g(z)=z$ on the strip where $0<\Re (z)<1$ and let $f(z)=1$ and $g(z)=0$ on the strip $2<\Re (z)<3$.