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A real valued function on the complex plane taking a complex number to its real part is an Open map,

just a hint please, first of all is it a continuous map? I did one problem in complex analysis which says if an analytic function takes only real values on $Y$ axis then it is a constant map which was clear to me by applying open map theorem as its image was closed set. what is the difference here?I know my $f$ is not given analytic.

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Recall that a map $f\colon X\rightarrow Y$ is open if and only if the image of basis elements of the topology on $X$ are open in $Y$. There's a very natural basis you can place on the complex plane. – Dan Rust Jan 29 '13 at 12:26
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You're really looking at the map $f(x,y) = x$ from $\mathbb{R}^2$ to $\mathbb{R}$ if I understand your question correctly.

This is an open mapping: If $U \subset \mathbb{R^2}$ is open and $(x,y) \in U$, then a small disc of radius $r$ centered at $(x,y)$ will be contained in $U$. Hence a small interval of length $2r$, centered at $x$ will be contained in $f(U)$.

(Note that this map is not open if you think of it as $\mathbb{R}^2 \ni (x,y) \mapsto (x,0) \in \mathbb{R}^2$.)

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What you want to show is that for any $U \subset \mathbb{R}$ that is open, its image (in $\mathbb{R}$) is open too.

So take any $x \in f(U)$. Our claim is now that there exists an $\epsilon >0$ s.t. $\forall x' \in (x-\epsilon, x+\epsilon): \exists w \in U s.t. f(w) = x'$.

Let $v \in U s.t. f(v) = x$, which must exist. We know (as $U$ is open) that $\exists \delta >0 s.t. \forall v' \in B(v,\delta): v'\in U$.

Finally set $\epsilon = \delta$ and take any $x' \in (x-\epsilon, x+\epsilon)$. One can see that $w':=(x',IM(x))\in B(v,\delta) \subset U$ and $f(w') = x'$ which proves the statement.

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