# Every point in a metric space has at least a neighborhood?

I was reading about topology and I came across this statement:

Every point $x$ in metric space $(X,d)$ has a neighborhood, which a neighborhood of $x$ (denoted $N(x)$) is defined as there exists $\delta >0$ such that the open ball $B_{\delta} (x) \subset N(x)$

The proof just says that take $N(x)= X$ as your neighborhood. Anybody can shed some light on how you can take $X$ be your neighborhood and ensure that every open ball around $x$ will lie completely in $X$?

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What is your definition of $B_\delta(x)$? I have difficulty imagining any that doesn't make $B_\delta(x)$ a subset of $X$ by definition. –  Henning Makholm Oct 4 '12 at 23:28
yes, points in $X$ whose distances are less than $\delta$ away from $x$ –  Daniel Oct 4 '12 at 23:29

## 2 Answers

Note that for any $\delta>0$ we have that $B_\delta(x)\subseteq X$. Therefore $X$ itself is a neighbourhood of any point $x\in X$.

Recall that $B_\delta(x)=\{u\in X\mid d(x,u)<\delta\}$. It follows immediately that $B_\delta(x)\subseteq X$.

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In the space $X$ everything is inside $X$, it is necessarily open, and contains $x$.

But you can also consider the open ball of radius 1 around $x$.

Anyway, this $N(x)$ in our class denoted the set of all neighborhoods of $x$: $$N(x):=\{ U\subseteq X \mid \exists\delta>0:B_\delta(x)\subseteq U\}$$

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Seriously? This is the second time already! :-) (I hate using my girlfriend's laptop...) –  Asaf Karagila Oct 4 '12 at 23:31
by the book I am using the $N(x)$ is just simply a neighborhood. According to my book your definition would be called something like a complete system of neighborhoods around $x$ –  Daniel Oct 4 '12 at 23:35