# 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? Mar 14, 2020 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. Mar 14, 2020 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. Mar 14, 2020 at 21:58
• Using a local base is not necessary. Just the open ball definiton of interior. Mar 14, 2020 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. Mar 14, 2020 at 22:15

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 Mar 14, 2020 at 22:19
• @rudinsimons12 yes, remove the local base bit. A picture gives the idea for a rigorous argument. Mar 14, 2020 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. $$\;\;\;\;\;\;\blacksquare\;$$