Proving set $(0,1)\times\{0\} \subseteq\mathbb{R^2}$ is not open.

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.

• Wouldn't it be easier to show the complement of that set is not closed? – DonAntonio Mar 14 '20 at 21:49
• @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. – rudinsimons12 Mar 14 '20 at 21:52
• 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. – zugzug Mar 14 '20 at 21:58
• Using a local base is not necessary. Just the open ball definiton of interior. – Henno Brandsma Mar 14 '20 at 21:59
• 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. – fleablood Mar 14 '20 at 22:15

2 Answers

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.

• 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 – rudinsimons12 Mar 14 '20 at 22:19
• @rudinsimons12 yes, remove the local base bit. A picture gives the idea for a rigorous argument. – Henno Brandsma Mar 14 '20 at 22:20

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. $$\;\;\;\;\;\;\blacksquare\;$$