# Proving $[0,1]\times\{0\} \subseteq\mathbb{R^2}$ is closed.

Can I please receive help in proving/solving the problem below? My proof is incomplete and I'm confused about how to prove this. $$\def\R{{\mathbb R}} \def\Rhat{{\widehat{\R}}}$$

So we know $$[0,1]$$ is closed in $$\R$$. I wish to prove, when considered as a subset of $$\R^2$$, that is, as a line segment on the $$x$$-axis in the plane, it is also closed. Specifically, I want to show that the set $$[0,1]\times\{0\} \subseteq\R^2$$ is closed.

$$\textbf{Solution:}$$ Let $$S = [0,1]\times\{0\}$$ and let's consider a point $$x=(a_1, b_1) \notin [0,1]\times\{0\}$$. Then, either $$a_1 \notin [0,1]$$ or $$b_1 \ne 0$$.

$$\textbf{Case I:}$$ $$a_1 \notin [0,1]$$

Let $$a_1 > 1$$ and similarly for $$a_1 < 1$$. Then choose $$\epsilon = \frac{a_1 -1}{2} > 0.$$ Then the neighborhood $$N_{\epsilon}(x)$$ does not contain any point of $$[0,1]\times\{0\}$$.

Therefore, $$x$$ is not a limit point of $$[0,1]\times\{0\}$$. Thus, $$[0,1]\times\{0\}$$ contains all it's limit points.

$$\textbf{Case II:}$$ If $$b_1 \ne 0$$. Then $$|b_1| > 0$$. Let's choose $$\frac{\epsilon}{2}>0.$$ Then the $$\epsilon$$-neighborhood $$N_{\epsilon} (x)$$ of $$x$$ contains no point of $$[0,1]\times\{0\}$$.

Hence, from Case I and Case II, we conclude that $$[0,1]\times\{0\}$$ contains all of its limit points. Hence $$[0,1]\times\{0\}$$ is a closed subset of $$\R^2$$. We conclude $$S^c$$ is open, because we found a neighborhood around every point, not in $$S^c$$ that is contained entirely in $$S^c$$. Since the complement of every open set is closed, $$S$$ is closed.

• You could just show that in the product topology a product of closed sets is closed. Mar 21 '20 at 3:33

It looks complete to me. You may use the fact that the complement of a closed set is open and vice versa. What your demonstration shows that the complement of S=[0,1]x{0} is open, so S is closed.

• Thank you for the feedback and help, herb! Mar 21 '20 at 3:15

Your proof is complete. Actually, you can also prove directly. Consider a sequence $$\{a_n\}_{n=1}^∞\subset S$$, then each $$a_n$$ can be written as $$(x_i,y_i)$$, $$x_i\in [0,1]$$ and $$y_i=0$$. If $$a_n \to a$$ as $$n\to ∞$$, and $$a=(x,y)$$, then we have $$y=0$$ since $$y_i=0$$, and $$x\in [0,1]$$ since $$x_i\in [0,1]$$. Consequently we have $$a\in S$$. In this way we also prove that S is closed.

• Thank you, Samuel! Mar 21 '20 at 18:32

Small point : from case $$1$$ and case $$2$$, what you conclude is that $$S^c$$ is open, because you've found a neighbourhood around every point not in $$S^c$$ that is contained entirely in $$S^c$$. Since the complement of every open set is closed, $$S$$ is closed.

Your proof of the fact that $$[0,1] \times \{0\}$$ is closed is then complete, and in fact very nicely written.

• Thank you for the feedback and help, астон! Mar 21 '20 at 3:15
• You are welcome, @rudinsimons12, do continue to post more often! I am sure you can afford to be more confident of your answers now. Mar 21 '20 at 3:19
• Thanks, I will. My main concern with this was showing what happens if $a_1 < 0$ in case I but also was not sure how to and if i was leaving anything else out. Mar 21 '20 at 3:31
• Not really. See, if $a_1 < 0$ the argument is so similar to the $a_1 > 1$ case, that if you wrote "and similarly for $a_1 < 0$" I would have given you full credit. It is true that you left the case out, but it was clear to me that you were aware of how to handle that case. Mar 21 '20 at 3:37
• Thank you, астон. I made the slight edit to the proof. Mar 21 '20 at 18:32