Show that $\overline{E_{1}\cup E_{2}}=\overline{E_1}\cup\overline{E_{2}}$.

Question

Show that $\overline{E_{1}\cup E_{2}}=\overline{E_1}\cup\overline{E_{2}}$.

Hi! I know ther is a better proof of this problem, but I wrote my solution, and I need verify this is righ or not, thanks!

My approach: $\rightarrow)$, Let $x\in\overline{E_{1}\cup E_{2}}$ , then we have two cases

1) $x\in E_{1}\cup E_{2}$, if and only if $x\in E_1$ or $x\in E_2$. Therefore, $x\in\overline{E_{1}}$ or $x\in\overline{E_{2}}$, then $x\in\overline{E_{1}}\cup \overline{E_{2}}$.

2) $x$ is limit point of $E_1\cup E_2$: Then there exist a sequence $(x_n)_{n}\subset E_1\cup E_2$ such that $x_n\to x$, when $n\to\infty$. Furtheremore, we have a finite union of set, then there exist a subsequence of $(x_n)$, said $(x_{n_{k}})_k$, which are in, for instance $E_1$, i.e., $(x_{n_{k}})\subset E_1$.

We can see, If we consider the open ball $B(x,\frac{1}{k})=\{x; 0<d(x,x_{n_{k}})<\frac{1}{k}\}$, then $x_{n_{k}}$ converges to $x$, because when $k\to\infty$, the distance between $x$ and $x_{n_{k}}$ is "very small". So we have that $x_{n_{k}}\to x$, $k\to\infty$ and $(x_{n_{k}})\subset E_1$, then $x$ is a limit point of $E_1$, this implies $x\in\overline{E_{1}}$ and this implies $x\in\overline{E}_{1}\cup\overline{E}_{2}$. Therefore, $$\overline{E_{1}\cup E_{2}}\subset\overline{E_1}\cup \overline{E_2}$$

$\leftarrow)$, Let $x\in \overline{E_1}\cup \overline{E_2}$, then $x\in\overline{E_1}$ or $x\in\overline{E_2}$. Without loss of generality, assume that $x\in\overline{E_1}$, then $x\in E_1$ or $x$ is a point limit of $E_1$. So again, we have two cases

1), If $x\in E_1\implies x\in E_1\cup E_2\implies x\in\overline{E_{1}\cup E_{2}}$.

2), If $x$ is a limit point of $E_{1}$, then $x$ is also a limit point of $E_{1}\cup E_2$, i.e., $x\in\overline{E_{1}\cup E_{2}}$.

So, we have that $\overline{E_1}\cup\overline{E_{2}}\subset\overline{E_{1}\cup E_{2}}$

Answer 1

Yes your answer is true. But some of your proof isn't need:

" We can see, If we consider the open ball $B(x,\frac{1}{k})=\{x; 0<d(x,x_{n_{k}})<\frac{1}{k}\}$, then $x_{n_{k}}$ converges to $x$, because when $k\to\infty$, the distance between $x$ and $x_{n_{k}}$ is "very small". "

Because if a sequence converges to a point $z_0$ then each subsequence of that converges to $z_0$ too.

Answer 2

You seem to be using the definition (one of the possible equivalent ones) of $\overline{A} = A \cup A'$, where $A'$ is the set of limit points of $A$.

And the fact (true in metric spaces): $x$ is a limit point of $A$ iff there is a sequence $a_n$ from $A$ that converges to $x$.

Under these assumptions the proof is correct, except for the vagueness that Mehr Gol also mentioned. In the first part we have a sequence $x_n$ from $E_1 \cup E_2$ that converges to $x$. Then indeed there is an $i\in \{1,2\}$ such that a whole subsequence $x_{n_k}$ lies in $E_i$ (the argument indeed hinges on the fact that $\mathbb{N}= \{n : x_n \in E_1\} \cup \{n : x_n \in E_2\}$, so at least one of the right hand sets must be infinite, and will give us that subsequence).

And as any subsequence of a convergent sequence converges to the same limit, $x_{n_k} \to x$ as well, and then the fact implies $x \in E'_i$ for the aforementioned $i$, and thus $x \in \overline{E_i}$ as well.

I think your proof would improve a lot by using the above ideas.