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If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, and $f : X → Y$ is a continuous map, is it true that for any open set $U ⊂ X$, the set $f(U)$ is open in Y ?

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    $\begingroup$ Usually, if $f$ satisfies that $f(U)$ is open for every $U$ open, then $f$ is called an open map. $\endgroup$
    – layman
    Nov 8, 2015 at 20:41

5 Answers 5

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This is not true in general.

Let $f: \Bbb R \to \Bbb R$ be defined as the constant function $f(x) = 2$. Then $f$ is continuous, but every open set is mapped to $\{2\}$, which is not open.

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No. The easiest counter example is $f(x) = c$. Then for all $U$, $f(U) = \{c\}$ is not open.

In general if $f$ is a real function, not injective, then open intervals can be found around a local max or min, the image of which will be an (at least) half-closed interval. Example: $f(x) = x^2$ has a minimum at $x = 0$ so for $a < 0 < b$, $f((a,b)) = [0, \max(a^2, b^2))$ is not open.

There are many more counter examples.

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No, but if you are looking for an interesting paper in this direction, consult:

Velleman, D. J. (1997). Characterizing continuity. American Mathematical Monthly, 318-322.

You can find a link to the article on JSTOR here.

A preview of the start:

enter image description here

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Another counter-example is for $U=(0,4 \pi)$ we have $\sin U = [-1,1]$

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Each function $f:(X,\tau_1)\to (X,\tau_2)$ where $\tau_1$ is Discrete Topology, is continous but there is no guarantee that $f$ be open

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