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I'm stuck with this one:

Show that $f:\mathbb{R^2}\to\mathbb{R}$, with $f(x,y)=x,$ is open and continuous, but not closed.

Here's what I've done so far:

$i)$ Open: Obviously the projection on the real axis of any open disk gives an open interval, so that's true.

$ii)$ Not closed: We can look at the graph of $g(x)=1/x$. It's a subset of $\mathbb{R^2}$, and closed with the usual topology in there, but it's projected into an open interval on $\mathbb{R}$.

What I cannot figure out is how to prove that this projection is continuous. That is, finding that open sets in $\mathbb{R^2}$ are mapped through $f^{-1}$ to open sets.

I'd appreciate any hint.

Thanks for your time.

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  • $\begingroup$ @quasi fixed, thanks. $\endgroup$
    – Relure
    Commented Oct 12, 2018 at 17:23
  • $\begingroup$ The projection function is just $f$ which is a polynomial, hence continuous. Applying $f$ to the points of the curve $y=1/x$ yields the domain of $g$, which is the set of all nonzero real numbers (which is not closed). $\endgroup$
    – quasi
    Commented Oct 12, 2018 at 17:24
  • $\begingroup$ Take a basis of open sets in $\Bbb R$ and show $f^{-1} (B)$ is open in $\Bbb R^2$, for instance if you take open intervals as the basis, then necessarily you have the pre-image eqaul to {$B, \Bbb R$}, which is open in the product topology. $\endgroup$
    – IntegrateThis
    Commented Oct 12, 2018 at 17:24
  • $\begingroup$ @quasi: is there way to show that polynomials are continuous which doesn't rely on the fact that the projection functions are continuous? $\endgroup$
    – Matthew Leingang
    Commented Oct 12, 2018 at 17:35
  • $\begingroup$ @Matthew Leingang: I guess not. But it's a question of what knowledge is considered already known. $\endgroup$
    – quasi
    Commented Oct 12, 2018 at 17:40

1 Answer 1

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You need to show that if $U$ is open in $\mathbb R$, then $f^{-1}(U)$ is open in $\mathbb R\times \mathbb R$.

But what is $f^{-1}(U)$? It's just $U\times \mathbb R$, which is clearly open in $\mathbb R\times \mathbb R$ (it is a basic open set in the product topology).

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  • $\begingroup$ Maybe that's a silly question, but didn't I find an open subset mapped onto a closed subset (on the non-closeness example)? Would't that contradict the definition of continuous? $\endgroup$
    – Relure
    Commented Oct 12, 2018 at 17:26
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    $\begingroup$ I don't see that, but it doesn't matter if an open set is mapped onto a closed set. Continuity only requires that the preimage $f^{-1}(U)$ of every open set $U$ is open. It doesn't matter whether or not the image $f(W)$ of an open set $W$ is open, or closed, or neither, or both. Note also: a closed set can also be open (this happens precisely when the set is a component of the space). $\endgroup$
    – MPW
    Commented Oct 12, 2018 at 17:33
  • $\begingroup$ @Relure. If $C$ is the graph of $f(x) = 1/x$ (closed), then $f(C) = \mathbb{R}\setminus\{0\}$ (open), but $f^{-1}(f(C)) = (\mathbb{R}\setminus\{0\})\times\mathbb{R}$ (still open). $\endgroup$
    – Matthew Leingang
    Commented Oct 12, 2018 at 17:34

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