Accumulation points of $\{ 2^{-n} + 5^{-m} : n,m \geq 1 \}$ 
Determine all of the accumulation points of the following sets in $\mathbb{R}^1$ and decide whether the sets are open or closed or neither.

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*The set $X$ of all numbers of the form $2^{-n} + 5^{-m}$ where $m,n = 1,2,\ldots$.


Claim: The set of accumulation points is the following:
$$\left\{\frac{1}{2^n}: n \in \mathbb{N} \right\} \cup \left\{\frac{1}{5^n}: n \in \mathbb{N}\right\} \cup \{0\}.$$
Let $n \in \Bbb{N}$ and $r>0$, and consider the open neighbourhood $B\left(\frac{1}{2^n},r\right) = \left(\frac{1}{2^n} - r,\frac{1}{2^n} + r\right)$.
For the given $r$ we can find $m$ large enough such that $5^{-m} < r \implies \frac{1}{2^n} + \frac{1}{5^m} < \frac{1}{2^n} + r$, and $\frac{1}{2^n} - \frac{1}{5^m} > \frac{1}{2^n} - r$; therefore $5^{-m} + 2^{-n} \in B\left(\frac{1}{2^n},r\right)$.
A similar argument shows that each element of $\{5^{-m}: m \in \mathbb{N}\} \cup \{0\}$ is an accumulation point of $X$.
Now consider $y \in X$. $B(y,r)$ isn't a subset of $X$ since for all $r > 0$, $B(y,r) \subset (\mathbb{R} - \mathbb{Q}) \not\subset \Bbb{Q}$, hence $B(y,r) \not\subset X$.
 A: The last argument is not correct.
For the missing argumentthat no other limit point exists:
The pair $1/2^{n}+1/5^{m}$ is determined by $(n,m)$. If a sequence approaches $x$, either $n$ or $m$ would go to $\infty$, which means the limit must conform to the expected forms.
A: Your argument for the fact that $1/2^n$ is an accumulation point works, I think. 
Have you tried justifying that $\{1/2^n\} \cup \{1/5^m\} \cup \{0\}$ are all the accumulation points?
I agree with you that $B(y,r)=(y-r,y+r)$ is not a subset of $X$, but your argument doesn't really justify it, because $y$, a rational number, is in $B(y,r)$, so $B(y,r) \not\subseteq (\Bbb{R}\setminus\Bbb{Q})$ .
However, to prove that $B(y,r) \not\subseteq X$ you don't really need to show that it consists solely of irrational numbers; it is enough to explain why it contains any irrational numbers at all. This follows from the fact that the set of irrational numbers is dense in $\Bbb{R}$, so every open interval contains irrational numbers. Thus, $B(y,r) \not\subseteq \Bbb{Q}$, and since $X \subseteq \Bbb{Q}$, this means that $B(y,r) \not\subseteq X$. 
