Just want to change Rudin theorem 4.8 a bit and see if this works.
The original theorem is ... $f$ is continuous iff $f^{-1}(V) $ is open in $X$ for every open set $V$ in $Y$.
If I change the inverse image to image, is this still true?
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Do you mean the criterion in your title, as opposed to Rudin's formulation? – Bernard Jul 05 '15 at 09:30
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It doesn't work. – Jul 05 '15 at 09:32
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@BolzWeir edited – maj Jul 05 '15 at 09:34
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@Bernard Yes you're right – maj Jul 05 '15 at 09:34
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I knew what you were asking about before the edit. It isn't true in general. – Jul 05 '15 at 09:35
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So what's the problem here? Can you briefly explain? @BolzWeir – maj Jul 05 '15 at 09:36
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1See: http://math.stackexchange.com/questions/75589/open-maps-which-are-not-continuous – Jul 05 '15 at 09:39
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A function $f:X\to Y$ is continuous if and only if $f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$. (This is sometimes taken to be the definition.) A function $f:X\to Y$ is open if and only if $f(V)$ is open in $Y$ for any open set $V\subset X$. (This is the definition.) These two are not the same.
Joonas Ilmavirta
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This is false. The property you mention for $f$ is called an open map:
(i) An open map can be non-continuous: suppose $Y$ is a discrete topological space. Then any map from $X$ to $Y$ is open, but can be non-continuous.
(ii) A continuous map is not necessarily open: if $Y$ is a non-open subset of $X$, with the induced topology, the canonical injection of $$ in $X$ is continuous, by definition of the induced topology, but the image of $Y$ is not open.
Bernard
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@maj $f(x)=x^2$ is a continuous map from $\mathbb{R}$ to $\mathbb{R}$. It is not open. – Jul 05 '15 at 17:11