Prove that the intersection of all neighborhoods of $x$ is $\{x\}$ I have to show that the intersection of the family of all neighborhoods of a point $x$ is $({x})$.
[$(x)$ is a singleton set]
My approach: Suppose there are n number of neighborhoods of $x$.
Let them be $y_1$,$y_2$,....$y_n$. Also $\epsilon{_n}<\epsilon_{n-1}<...<\epsilon_{1}$
Then by definition of neighborhoods, we have, $x\in (x-\epsilon{_1},x+\epsilon{_1})\subset y_1$   *
$x\in (x-\epsilon{_2},x+\epsilon{_2})\subset y_2$
.
.
.
Similarly, $x\in (x-\epsilon{_n},x+\epsilon{_n})\subset y_n$   **
Then by * and **, and taking intersection of all neighborhoods, we get their intersection $(x)$
Is this approach correct? Thanks in advance!
 A: Let $I$ be the intersection of all neighborhoods of $x$. I think it’s clear that $\{x\} \subset I$. Now we have to show that there are no other points (other than $x$) in $I$. So, suppose $y \ne x$. We need to show that $y \not\in I$. To do this, you have to find a neighborhood of $x$ that does not contain $y$. Can you do that?
Hint: it might be useful to think about the point $z$ that’s half-way between $x$ and $y$.
Added later: Since you already accepted another answer, I’ll complete the details of mine.
Let $\delta = \tfrac12|y-x|$. Then $\delta >0$ and so the open interval $(x-\delta, x+\delta)$ is a neighborhood of $x$ that does not contain $y$, and this shows that $y\not\in I$.
A: To restate your definition,

A set $N$ is called a neighborhood of a point $p$, if there exists an open interval $I$ containing $p$, and contained in the set, i.e. $p\in I \subset N$.


Firstly, your approach considers only a finite number of neighborhoods. There are actually infinitely many of them. Since we do not know how many neighborhoods there are (and what they look like), let us denote the set of all neighborhoods of some point $x$ by $\{N_i\}_{i\in I}$, where $I$ is some index set.
Strategy:

*

*Show $\{x\} \subset \bigcap_{i\in I} N_i$.

*Show $\{x\} \supset \bigcap_{i\in I} N_i$.

*Conclude $\{x\} = \bigcap_{i\in I} N_i$.


By definition, it is clear that $x\in N_i$ for every $i$, so $x\in \bigcap_{i\in I} N_i$. In other words, $\{x\} \subset \bigcap_{i\in I} N_i$.
Observe that for every natural number $n\in\mathbb N$, $(x-\frac1n, x+\frac1n)$ is a neighborhood of $x$. We have
$$\bigcap_{i\in I} N_i \subset \left(x-\frac1n, x+\frac1n\right)$$
In fact, since this holds true for every $n\in\mathbb N$, we get
$$\bigcap_{i\in I} N_i \subset \bigcap_{n=1}^\infty\left(x-\frac1n, x+\frac1n\right)$$
It is a good (and standard) exercise to show that $$\bigcap_{n=1}^\infty\left(x-\frac1n, x+\frac1n\right) = \{x\}$$
Since $x$ was arbitrarily chosen, the result holds for every $x\in\mathbb R$. We have the conclusion.
A: First
Suppose $x\in\mathbb R,\epsilon_n \searrow 0$, then $\{x\}=\bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)$
Proof: On the one hand, $x\in(x-\epsilon_n,x+\epsilon_n)\forall n$, therefore  $x\in\bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)$. On the other hand, suppose $\exists y,\vert y-x\vert >0, y\in \bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)$. Considering that $\epsilon_n\searrow 0$, one can find $N>0$ such that $\epsilon_n < \vert y-x \vert$ whenever $n>N$, then $y\notin (x-\epsilon_{n},x+\epsilon_n)$ whenever $n>N$. Contradiction.
In conclusion $\{x\}=\bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)$
By definition, $\{(x-\epsilon_n,x+\epsilon_n)\}_{n=1}^\infty$ are neighborhoods of $x$, then intersection of all neighborhoods of $x$:$\bigcap\limits_{x\in I\subset N_i}N_i$ is subset of $\bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)$
Then $\{x\}\subset\bigcap\limits_{x\in I\subset N_i}N_i\subset\bigcap\limits_{n=1}^\infty(x-\epsilon_n,x+\epsilon_n)=\{x\}$, i.e. $\bigcap\limits_{x\in I\subset N_i}N_i=\{x\}$
