Prove/disprove if $f:R \to R$ is continuous and $U\subset R$ is open, then $f(U)\subset R$ is open

Asked Sep 27 '21 at 22:52

Active Sep 28 '21 at 08:22

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Given $f:X\to Y$ is a function and $U \subset X$, the image of $U$ under $f$ is $f(U)=\{y\in Y: y = f(x)$ for some $x\in U\}$ Prove/disprove if $f:R \to R$ is continuous and $U\subset R$ is open, then $f(U)\subset R$ is open

I mean this intuitively makes sense. If there's an open set of U, then any function that is done upon the elements of U must also be open as well. But I don't know any direction I could go from here. I saw this post which is kind of similar but doesn't answer my question and this was a failed proof. Any suggestions?

edited Sep 28 '21 at 08:22

Paul Frost

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asked Sep 27 '21 at 22:52

John Rawls

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Have you considered a function like $y=\cos(x)$ on $(-\pi/2,5\pi/2)?$ – Matthew H. Sep 27 '21 at 22:56
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or even more simply, $f(x)=x^2$? – Greg Martin Sep 27 '21 at 22:57
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or yet more simply, a constant function such as $f(x) \equiv 0$ – Daniel Schepler Sep 27 '21 at 23:24
This property is akin to having (A suitably restricted if f is not a bijection) the inverse being continuous, and thus the function being called a homeomorphism – Alan Sep 28 '21 at 00:40
What is $R$? The reals? – Paul Frost Sep 28 '21 at 08:21
This is a duplicate question ... see https://math.stackexchange.com/q/1519620/145325 the answers point to counterexamples for the case when the domain and range is the set of reals – TryingHardToBecomeAGoodPrSlvr Sep 28 '21 at 08:29

Prove/disprove if $f:R \to R$ is continuous and $U\subset R$ is open, then $f(U)\subset R$ is open

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