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Can I please get some help/feedback on my proof? Thank you. $\def\R{{\mathbb R}} \def\Rhat{{\widehat{\R}}} \def\N{{\mathbb N}}$

$(0,1)$ is open in $\R$. I will prove that, when considered as a subset of $\R^2$, that is, as a line segment on the $x$-axis in the plane, it is not open. Specifically, I will show that the set $(0,1)\times\{0\} \subseteq\R^2$ is not open.

$\textbf{Solution:}$ Consider the set $(0,1) \times \{0\} \subset \R^2.$ A set $S$ is called open if every point of the set $S$ is an interior point, that is, for every point $x\in S$, there exist an open set $V$, such that $x\in V \subset S.$

Pick the point $p = (\frac{1}{2}, 0)$ inside $(0,1) \times \{0\}$. In $\R^2$, the open balls form a basis for the topology of $\R^2$, meaning every point $x$ in $\R^2$, we can find an open ball containing it and if an open set $U$ contains $x$, there exists an open ball centered at $x$ such that $x\in B \subset U$. So, if we can show there do not exist any open ball centered at $p = (\frac{1}{2}, 0)$ contained in $(0,1) \times \{0\}$, we will be done.

Now, we will show why no open ball sits inside $(0,1) \times \{0\}$ by supposing it is, that is, there is an open ball $B(p,r)$, for some $r>0$, in the Euclidean metric on $\Bbb R^2$ such that $$B(p,r) \subseteq (0,1) \times \{0\}\tag{1}$$

But $q=(\frac12, \frac{r}{2})$ obeys $d(p,q)=\frac{r}{2}< r$, so that $q \in B(p,r)$ but as $\frac{r}{2} \neq 0$, $q \notin (0,1) \times \{0\}$. This contradicts our supposed inclusion $(1)$. So $p$ is not an interior point of $(0,1) \times \{0\}$ and $(0,1) \times \{0\}$ is not open.

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  • $\begingroup$ Wouldn't it be easier to show the complement of that set is not closed? $\endgroup$ – DonAntonio Mar 14 '20 at 21:49
  • $\begingroup$ @DonAntonio I think but I am not able to, I posted it earlier and my proof for that was far off apparently so I hope this is better but I am not entirely sure. $\endgroup$ – rudinsimons12 Mar 14 '20 at 21:52
  • $\begingroup$ That's a very thorough answer. Do you really need all of the stuff about bases? I've always thought the standard definition of an interior point is that there exists a neighborhood that is contained within the set. In your definition, you say there exists an open set, which is true, but more complicated than just sticking with neighborhoods. $\endgroup$ – zugzug Mar 14 '20 at 21:58
  • $\begingroup$ Using a local base is not necessary. Just the open ball definiton of interior. $\endgroup$ – Henno Brandsma Mar 14 '20 at 21:59
  • $\begingroup$ It seems like you did a lot of talking to say basic concepts, which is fine, always err on more than less when you are learning, but then the entire gyst is this sentence " Now, for any n∈N, the open ball B1/n(p) do not lie inside the set (0,1)×{0}" which is stated as fact without any verification. $\endgroup$ – fleablood Mar 14 '20 at 22:15
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The idea is fine: show that e.g. $p=(\frac12,0)$ is not an interior point of $(0,1) \times \{0\}$. But you do not show why no open ball sits inside $(0,1) \times \{0\}$, you need to fill that gap (e.g. a picture is not a proof!)

So suppose it is, so there is an open ball $B(p,r)$, for some $r>0$, in the Euclidean metric on $\Bbb R^2$ such that $$B(p,r) \subseteq (0,1) \times \{0\}\tag{1}$$

But $q=(\frac12, \frac{r}{2})$ obeys $d(p,q)=\frac{r}{2}< r$, so that $q \in B(p,r)$ but as $\frac{r}{2} \neq 0$, $q \notin (0,1) \times \{0\}$. This contradicts our supposed inclusion $(1)$. So $p$ is not an interior point of $(0,1) \times \{0\}$ and $(0,1) \times \{0\}$ is not open.

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  • $\begingroup$ Thank you Henno for the feedback and help. Do you think I should just remove the local basis claim? I made an edit to my solution to include yours. I am trying to reread this, it is easier to prove these sort of things by picture but proof by picture is not a proof lol $\endgroup$ – rudinsimons12 Mar 14 '20 at 22:19
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    $\begingroup$ @rudinsimons12 yes, remove the local base bit. A picture gives the idea for a rigorous argument. $\endgroup$ – Henno Brandsma Mar 14 '20 at 22:20
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Another idea: take for example the sequence

$$\left\{\left(\frac12\,,\,\frac1n\right)\right\}_{n\in\Bbb N}$$

Check that the above sequence is not in the set $\;E:=(0,1)\times\{0\}\;$, so it belongs to $\;E^c=\Bbb R^2\setminus E\;$ . The sequence is clearly convergent, so if $\;E^c\;$ is closed it must contain this sequence's limit, yet

$$\lim_{n\to\infty}\left(\frac12\,,\,\frac1n\right)=\left(\frac12,\,0\right)\in E$$

which means $\;E^c\;$ is not closed and thus $\;(E^c)^c=E\;$ is not open. $\;\;\;\;\;\;\blacksquare\;$

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