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.
Best Answer
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.