Writing a singleton as a countable intersection Let $x \in \mathbb{R}$. Then I am almost sure, that $$\{x\} = \bigcap_{n \in \mathbb{N}}(x - 1/n,x]$$
Any ideas how to prove this?
 A: Obviously $x\in(x-\frac{1}{n},x]$ for all $n$. So $x\in\bigcap_{n \in \mathbb{N}}(x - 1/n,x]$. Also let $y\neq x$. Then we can find $n_0\in \mathbb{N}$ such that $x-y>\frac{1}{n_0}$. Then $y\not\in(x-\frac{1}{n},x]$ for all $n\geq n_0$. Hence $y\not\in\bigcap_{n \in \mathbb{N}}(x - 1/n,x]$.
A: Suppose $a$ is in $\bigcap_{n\in\mathbb{N}}(x-1/n,x]$
Then $a$ is in $(x-1/n,x]$ for all $n\in\mathbb N$.
Let $\varepsilon >0$.  I claim that $a$ is within $\varepsilon$ of $x$.  Don't believe me?  Take $n$ large enough so that $1/n < \varepsilon$.  Then $a$ is in $(x-1/n, x] \subset (x-\varepsilon, x]$.
In other words, for all $\varepsilon >0$, there exists an $n$ such that $|x-a| < \varepsilon$.  Since $x$ and $a$ are fixed, I'll go out on a limb to say that $x=a$.
The reverse is trivial.
A: Note that $x\in (x-1/n, x]$ for every $n\in \mathbb{N}$. On the other hand, given $y\in \bigcap_{n \in \mathbb{N}}(x - 1/n,x]$, we have $y\in (x - 1/n,x]$ for every $n\ge 1$. So $x-1/n< y\le x$. If we make $n\rightarrow \infty$, by sandwich theorem we get $x\le y\le x$. So $y=x$ and hence, $$\{x\} = \bigcap_{n \in \mathbb{N}}(x - 1/n,x].$$
