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Suppose $X\subset \mathbb R^2$ with the subset topology and $U\subset\mathbb R^2$ an open subset with the subset topology, if $X\cong U$ and we deduce $X\subset \mathbb R^2$ is also open?

This is not true for general topological space, for example let $S=\{1,2\}$ with the topology \begin{equation} \tau_S=\big\{\emptyset,\{1,2\},\{1\}\big\} \end{equation} then \begin{equation} \{2\}\cong\{1\} \end{equation} but $\{1\}\subset S$ is open and $\{2\}\subset S$ is not open.

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    $\begingroup$ Yes: an injective continuous map from an open set of $\Bbb R^n$ to $\Bbb R^n$ is open. This is known as invariance of domain theorem; it's a result that uses some algebraic topology (or some other non-elementary tool). $\endgroup$
    – user562983
    Commented Jul 13, 2018 at 15:20
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    $\begingroup$ @SaucyO'Path. Yes. It's one of those "obvious" theorems that are actually hard to prove. (Like the Jordan Curve Theorem.) $\endgroup$ Commented Jul 13, 2018 at 20:25
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    $\begingroup$ No need for such an exotic counterexample. In the space $S=[0,\infty)$ (topologized as usual), the set $X=[1,\infty)$ is homoemorphic to the whole space, but is not open. $\endgroup$
    – bof
    Commented Jul 13, 2018 at 22:30
  • $\begingroup$ @DanielWainfleet Regarding the proof, the wikipedia page en.wikipedia.org/wiki/Invariance_of_domain says that it uses "notably the Brouwer fixed point theorem.". If one takes this fixed point theorem for granted, do you know if the proof is 'reasonably elementary' for someone who doesn't know any algebraic topology? $\endgroup$ Commented Jul 17, 2018 at 17:52
  • $\begingroup$ @CalvinKhor . I'm a senior & there are some proofs that I haven't thought about for a long time, so I dk. What I do recall is that proving the Brouwer fixed point theorem in 2 or more dimensions is not obvious. $\endgroup$ Commented Jul 17, 2018 at 18:58

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Yes, this is true for any $\mathbb{R}^n$ as a consequence of the invariance of domain theorem. If

$$f:U\to X$$

is a homeomorphism then we have an injective continuous map $g:U\to\mathbb{R}^n$, $g(x)=f(x)$ which has to be open. Thus the image, being $X$ is open.

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