$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Note that for any point $x\in ℝ\setminus S$, what we will do is that we will assume there was a sequence that converges to $x$ then if it can be shown that there exists an epsilon ball which is completely disjoint from $S$ then $x$ certainly cannot be a limit point creating a contradiction. If this can be done for all points in $ℝ\setminus S$ then it will have show that $S$ already contains all of its limit points.

Formally if $x\in ℝ\setminus S$ then if $x\notin [0,1]$ and if $x>1$ we can use $ϵ=\frac{x-1}{2}$ and if $x<0$ then we can use $ϵ=\frac{\left|x\right|}{2}$. Otherwise $x\in [0,1]$, but we also know that $x\notin \{\frac{1}{n}:n\in {\mathbb{N}}_1\}\cup \left\{0\right\}$, so certainly $x\notin \{0,1\}$ and moreover what can be observed is that there exists a $k\in {\mathbb{N}}_1$ such that $x\in (\frac{1}{k},\frac{1}{k+1})$, thus by taking $ϵ=\frac{\min (x-\frac{1}{k+1},\frac{1}{k}-x)}{2}$, we can find an epsilon ball that completely excludes any points from $S$ so $x$ cannot be a limit point.

In order to do so we have to show that $A$ contains all of its limit points, let $\left(x_n\right):{\mathbb{N}}_1\to A$ be a sequence such that $\left(x_n\right)\to x$ let's prove that $x\in A$.

If $\{x_n:n\in {\mathbb{N}}_1\}$ is finite, then it converges to a point in $\{x_n:n\in {\mathbb{N}}_1\}\subseteq A$ as needed.

On the other hand if $\{x_n:n\in {\mathbb{N}}_1\}$ is infinite, then that shows that $A$ must be infinite as well, speficially that $\{a_n:n\in {\mathbb{N}}_1\}$ is infinite, moreover we can see that $\{x_n:n\in {\mathbb{N}}_1\}\subseteq A$, our goal will be to show that $x=a$ which can be done by showing that for any $ϵ\in {ℝ}^{+}$ we have that $\Vert x-a\Vert <ϵ$. To start things off let $ϵ\in {ℝ}^{+}$:

Given $\left(x_n\right)$ we can create a subsequence $\left(y_n\right)$ which is the same, but it omits any element such that $x_i=a$, firstly note that $\{y_n:n\in {\mathbb{N}}_1\}=\{x_n:n\in {\mathbb{N}}_1\}\setminus \left\{a\right\}$ and thus still has infinite size. By doing this see that for any $k\in {\mathbb{N}}_1$ we have that $y_k=a_{f\left(k\right)}$ where $f:{\mathbb{N}}_1\to {\mathbb{N}}_1$ is a function (not necessarily increasing), but must have an infinite image, otherwise $\{y_n:n\in {\mathbb{N}}_1\}$ would be finite.

Since $\left(x_n\right)\to x$ then we know that $\left(y_n\right)\to x$ since it's subsequence. Now using $\frac{ϵ}{2}$, then we get some $N^{\prime }\in {\mathbb{N}}_1$ such that for any $n\ge N^{\prime }$ we have that $\Vert y_n-x\Vert =\Vert a_{f\left(k\right)}-x\Vert <\frac{ϵ}{2}$

On the other hand we know that $\left(a_n\right)\to a$ and therefore for that same choice of epsilon we obtain some $N\in {\mathbb{N}}_1$ such that for any $n\ge N$ we have that $\Vert a_n-a\Vert <\frac{ϵ}{2}$.

Now since $f$ has infinite image, then we know that there is an $x\ge N$ such that $f\left(x\right)\ge N^{\prime }$ therefore we see that $$\Vert a_{f\left(x\right)}-a\Vert <\frac{ϵ}{2}\text{ and }\Vert a_{f\left(x\right)}-x\Vert <\frac{ϵ}{2}$$ by adding the two inequalities and moving left using the triangle inequality we obtain that $\Vert x-a\Vert <ϵ$ so that $x=a$, and thus $x\in A$ so $A$ is closed.

$\subseteq$ Let $p\in \overline{A}$, then if $p\in A$ we are done, as $\left\{p\right\}\subseteq B(p,ϵ)\cap A$ for any $ϵ$, on the other hand if $p\notin A$ then since $\overline{A}$ is the set of limit points of $A$ there exists a sequence $\left(x_n\right):{\mathbb{N}}_1\to A$ such that $\left(x_n\right)\to p$, now let $ϵ\in {ℝ}^{+}$ therefore there exists an $N\in {\mathbb{N}}_1$ such that for all $n\ge N$ we have $\Vert x_n-p\Vert <ϵ$ therefore $x_n\in B(p,ϵ)$ and since $x_n\in A$ then $\varnothing \ne \left\{x_n\right\}\subseteq B(p,ϵ)\cap A$ as needed.

$\supseteq$ On the other hand given a point $p$ from the right hand side, we want to show that $p\in \overline{A}$ which is that there is a sequence entirely contained in $A$ whose limit is $p$. This can be done as for any $k\in {\mathbb{N}}_1$ then since there exists some $ϵ\in {ℝ}^{+}$ such that $ϵ\le \frac{1}{k}$ then we can define a sequence $x_k$ such that $x_k\in B(p,\frac{1}{k})$, so that $\left(x_n\right)\to p$ as needed.

We want to prove that $L+M$ is closed, so in order to do that we have to show that it contains all of its limit points. So suppose that $\left(z_n\right):{\mathbb{N}}_1\to L+M$ is a sequence that converges to a point $z$ then we want to show that $z\in L+M$.

Note that for any $k\in {\mathbb{N}}_1$ there must exist some $l_k,m_k\in L,M$ respectively such that $z_k=l_k+m_k$. So we can consider $\left(m_n\right):{\mathbb{N}}_1\to M$, since $M$ is compact then there exists some $m_{\sigma \left(n\right)}:{\mathbb{N}}_1\to M$ such that $\left(m_{\sigma \left(n\right)}\right)\to m$ where $m\in M$.

Next consider the sequence $z_{\sigma \left(n\right)}:=l_{\sigma \left(n\right)}+m_{\sigma \left(n\right)}$, since $\left(z_n\right)$ converges then so does $\left(z_{\sigma \left(n\right)}\right)$, as just discussed $\left(m_{\sigma \left(n\right)}\right)$ converges, therefore $l_{\sigma \left(n\right)}$ converges, but $L$ is closed so it converges to a point $l\in L$.

Thus we can see that $\left(l_{\sigma \left(n\right)}\right)\to l$, $\left(m_{\sigma \left(n\right)}\right)\to m$ therefore we know that $\left(z_{\sigma \left(n\right)}\right)\to m+l$. Recall that we assumed that $\left(z_n\right)\to z$, and we know that any subsequence also converges to $z$ which implies that $z=m+l$ thus since $m\in M$ and $l\in L$ we know that $z\in M+L$ as needed.

Let $J:=\{n+\frac{1}{n}:n\in \mathbb{N}\}$ represent the set on the left hand side. To see why it is closed, evaluate the few first terms to see $1+1,2+\frac{1}{2},3+\frac{1}{3},\dots$, we'll show that for any $x\ne y\in J,|x-y|\ge \frac{1}{2}$, we'll first cover the case $x=1+1,y=2+\frac{1}{2}$ (wlog) and so $|x-y|=|2-2.5|=\frac{1}{2}\ge \frac{1}{2}$ as needed, now suppose that $x=l+\frac{1}{l}$ and $y=m+\frac{1}{m}$ such that $l,m\ge 2$ where without loss of generality we have $l<m$ therefore we know that $\frac{1}{m}<\frac{1}{l}$, from this we conclude that $$|x-y|=|m-l+(\frac{1}{l}-\frac{1}{m})|\ge 1\ge \frac{1}{2}$$ thus for any $x\in ℝ\setminus J$ by using $ϵ=\frac{1}{4}$ we can get a contradiction if such a limit were to exist, thus $J$ is closed.

The right hand side (which we will denote by $K$) is closed as well for the even simpler $ϵ=\frac{1}{2}$

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But their sum is certainly not closed, because $0\notin J+K$ but $\{\frac{1}{n}:n\in {\mathbb{N}}_1\}\subseteq J+K$, this means that $0$ is a limit point which is not contained in $J+K$, so $J+K$ is not closed.

Suppose for the sake of contradiction that $\left(a_n\right)$ does not converge to $a$, thus there exists some $ϵ\in {ℝ}^{+}$ such that for any $N\in {\mathbb{N}}_1$ there exists an $n\ge N$ and $\Vert a_n-a\Vert \ge ϵ$.

Thus for $N=1$ we obtain some $n_1$ such that $\Vert a_{n_1}-a\Vert >ϵ$, then recursively, by setting $N=n_i$ we obtain some $n_{i+1}$ for $i\in {\mathbb{N}}_1$ such that $\Vert a_{n_{i+1}}-a\Vert \ge ϵ$, by doing this we obtain that $n_k\le n_{k+1}$ for all $k\in {\mathbb{N}}_1$, so that we've constructed a subsequence $a_{n_k}$ such that for any $k\in {\mathbb{N}}_1$ we have that $$\Vert a_{n_k}-a\Vert \ge ϵ$$

But as per assumption since $\left(a_{n_k}\right)$ is a subsequence then it has a sub-subsequence that converges to $a$, but this is clearly impossible since all elements of this subsequence stay at least $ϵ$ away from $a$. Therefore it must be that $\left(a_n\right)\to a$

Suppose the former, but for the sake of contradiction assume that the latter is false, which is that every convergent subsequence converges to the same value, let's denote this value as $L\in {ℝ}^n$

Let $\left(x_{f\left(n\right)}\right)$ be a subsequence, since $K$ is compact, then it is bounded, thus since $x_n$ is a value in $K$ then $\left(x_n\right)$ is a bounded sequence in ${ℝ}^n$ therefore by Bolzano-Weierstrass, we see that $x_{f\left(n\right)}$ converges, and thus by our previous paragraph $\left(x_{f\left(n\right)}\right)\to L$, thus for any subsequence of $\left(x_{f\left(n\right)}\right)$ it also converges to $L$ therefore $\left(x_n\right)\to L$ which is a contradiction.

Since we've reached a contradiction, then we know that our assumption was false so that there are two subsequence of $\left(x_n\right)$ that converge to different values.

We want to show that $A\cap U$ is dense in $U$ which is to say that $U\subseteq \overline{A\cap U}$, thus let $u\in U$ we want to construct a sequence $\left(x_n\right):{\mathbb{N}}_1\to A\cap U$ such that $\left(x_n\right)\to u$.

Recall that $A$ was dense in ${ℝ}^n$ which means that ${ℝ}^n\subseteq A$ in other words $A={ℝ}^n$, since $u\in U\subseteq {ℝ}^n$ then $u\in {ℝ}^n$ and thus there exists a sequence $\left(a_n\right):{\mathbb{N}}_1\to A$ such that $\left(a_n\right)\to u$.

On the other hand we know that $U$ is open and thus there is some $ϵ\in {ℝ}^{+}$ such that $B(u,ϵ)\subseteq U$. Using this $ϵ$ in the fact that $\left(a_n\right)\to u$ we see that there is some $N\in {\mathbb{N}}_1$ such that for all $n\ge N$ we have that $\Vert a_n-u\Vert <ϵ⟺a_n\in B(u,ϵ)$.

With all this in mind we construct $x_n=a_{N+n}$ so that we have $x_n\in A\cap B(u,ϵ)\subseteq A\cap U$ moreover, since $\left(x_n\right)$ is a subsequence $\left(a_n\right)$ then we know that $\left(x_n\right)\to u$ as needed.

Since $x$ is a limit point of $A$ then we get some $\left(x_n\right):{\mathbb{N}}_1\to A$ such that $\left(x_n\right)\to x$. Now there are two cases, if $x\in A$ we are done, otherwise $x\notin A$ then it must be the case that for all $k\in {\mathbb{N}}_1$ we have that $x_k\ne x$ otherwise there would be some $j\in {\mathbb{N}}_1$ such that $x_j=x\notin A$ which is a contradiction, therefore by considering the identical sequence but just restricting it's range $\left(y_n\right):{\mathbb{N}}_1\to A\setminus \left\{x\right\}$ since $y_n=x_n$ for all $n\in {\mathbb{N}}_1$ then we know that $\left(y_n\right)\to x$ as needed.