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I suppose that $f:E\rightarrow F$ is continuous and surjective

To prove that $f$ i closed i must prove that the direct image of any closed set, is closed in $F$ .

As $f$ is continuous we have that $f(\overline{A})\subset \overline{f(A)}, ~\forall A\subset E$

But to obtain that it is closed i mut have that $\overline{f(A)}\subset f(\overline{A})~\forall A\subset E$

Can i prove this using the surjectivity of $f$ ?

Thank you

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    Injective or surjective? – 2017-02-19
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    surjective i'm sorry – 2017-02-19
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    HInt: Consider $t\to e^{it}$ from $[0,2\pi)$ to the unit circle. – 2017-02-19
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    What you are trying to prove is false even for maps of subsets of $R$. – 2017-02-19
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    the question is let $f$ continuous and surjective, prove that the image of any dense set in E is dense in $F$ , (this question is simple) after that the question is does $f$ is closed ? why ? – 2017-02-19
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    @MoisheCohen can you give me a simple example please – 2017-02-19
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    @zhw. this example is quite diffecult – 2017-02-19
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    No it isn't. It's just the map $t\to (\cos t, \sin t).$ – 2017-02-19
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    but $[0,2\pi)$ is not closed in $(\mathbb{R},|.|)$ we must take $[0,2\pi]$ and i don't know what we find $f([0,2\pi])$ – 2017-02-19
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    @Vrouvrou It does not matter that $[0,2\pi)$ is not closed. You put the subspace topology on the interval and consider some sequence that converges to $2\pi$ in it. – 2017-02-19
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    can we find a simple example without using sequces ? it is not on the chapter we study – 2017-02-19

2 Answers 2

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Consider the projection of real plane to X-axis ie. a continuous surjective mapping from $\mathbb R^2$ to $\mathbb R$. Now consider the image of the closed set {(x,y): xy=1}, it's $\mathbb R- ${0} which is not closed.

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Here is an example of a continuous surjective function $f:E\to F$ that is not closed and doesn't use sequences.

Take $E=F=\mathbb{R}$ and define $f$ by the following:

$$f =\begin{cases} \arctan(x)\sin(x) \text{ for } x\geq 0 \\ x\sin(x) \text{ for } x<0. \end{cases} $$ This is surjective and continuous. However observe that the image of $[0,\infty)$ is $(-\pi/2 ,\pi /2)$ which is open.

Edit: Now that I have graphed your suggested function(in red) verse my function(in blue), do you now understand why your suggestion is not surjective and my use of $\sin$? enter image description here

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    why you added $\sin(x)$ you can define $f$ as $$f=\begin{cases} arctan(x)~ \text{for}~ x\geq 0\\ x ~ \text{for}~ x<0\end{cases}$$ we obtain the same result no ? – 2017-02-20
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    That function is not surjective onto $\mathbb{R}$. I have $\sin$ because I was originally thinking in terms of sequences and the peaks of the $\sin$ function to construct a counterexample. Also, for a function $f:\mathbb{R}\to\mathbb{R}$, if we add the assumption of monotonicity (along of with continuity and surjectivity) then $f$ is closed. So we must have some sort of non-monotonic behaviour in this case. – 2017-02-20
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    i don't unserdand why it is not surjective, and we find the same result $f([0,+\infty[)=]-\frac{\pi}{2}, \frac{\pi}{2}[$ no ? – 2017-02-20
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    I suggest you try graphing the function you wrote. Note that it is bounded above by $\pi/2$ so not surjective and $f([0,\infty))= [0,\pi/2)$. – 2017-02-20
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    so how we get $-\frac{\pi}{2}$ when x\geq0 in your function – 2017-02-20
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    That's where the $\sin$ comes in. – 2017-02-20
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    i don't understanw we found this – 2017-02-20
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    See my update. If you are still confused then we should move this to a chat. – 2017-02-20
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53932/discussion-between-vrouvrou-and-leon-sot). – 2017-02-20