Is there an explicit way to show this using theorems in real/complex analysis, without using topology?
2026-03-26 04:29:33.1774499373
Re(z) is an open map
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Well obviously some topological notions are necessary; but a quick way to get at this is note that holomorphic functions $f(z)$ are open maps (the open mappig theorem), and the projection $\pi_{\Bbb R}$ onto the real axis is open (see this wikipedia page), so
$\Re(f) = \pi_{\Bbb R} \circ f \tag 1$
is open.
As pointed out by users 797616 Trash Failure in their comments to this answer, there is no need to invoke the openness of general holomorphic $f(z)$ since all that is asked for is the openness of $\Re(z)$; my enthusiasm for generalization got the better of me.