I keep seeing the closure of a set to be defined as follows:

$\textbf{Definition:}$ Let $(X, d)$ be a metric space. Also, let $A\subseteq X$.

Then, $x\in cl(A)$ iff $\forall \epsilon>0$, $B(x;\epsilon)\cap A\neq \phi$ $\tag1$

I personally use the definition at the top shown below which I tagged as $(2)$.

$\textbf{Question:}$ How are these definitions equivalent (i.e. (1) and (2))? Any help would be appreciated!

\begin{align} x\in cl(A) \leftrightarrow \forall \text{ neighborhood } N \text{ of } x, N\cap A\neq \phi \tag2 \end{align}

Also, I use the following definition of a neighborhood: $N$ is a $\textbf{neighborhood}$ of $a$ given a metric space $(X, d)$ iff $\exists \delta\in \mathbb{R}^{>0}$ ST $B(a;\delta)\subseteq N\subseteq X$.

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    $\begingroup$ I do not think they are equivalent. Example in $\mathbb R$ with standard metric, $S = (1,2)$, then $2 \in \mathrm {cl}(S)$, but none of neighborhood $B(2;r)$ is contained in $S$. $\endgroup$ – xbh Aug 29 '18 at 1:52
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    $\begingroup$ They are simply not equivalent. $\endgroup$ – Eduardo Longa Aug 29 '18 at 1:54
  • $\begingroup$ Okay. What definition does everybody use then? I am surprised they aren't equivalent. $\endgroup$ – W. G. Aug 29 '18 at 1:56
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    $\begingroup$ @W.G. $(1)$ seems a widely acceptable choice. If you like, you could change $B(x; \varepsilon)$ to "neighborhood" [although the latter might need some work to prove]. $\endgroup$ – xbh Aug 29 '18 at 2:01
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    $\begingroup$ In topology the definition of a neighborhood of a point $p $ in a space $S$ is a set $N\subset S$ such that $ p\in U\subset N$ for some open $U.$ In a metric space this is equivalent to the def'n you are using. $\endgroup$ – DanielWainfleet Aug 29 '18 at 16:17

According to your definition of neighborhood, every open ball centered at x is a neighborhood of x and every neighborhood of x contains an open ball centered at x.

Thus if every neighborhood of x intersects A then every open ball will intersect A and if every open ball centered at x intersects A, then every neighborhood of x intersects A because it contains an open ball centered at x.

Therefore these two definitions are equivalent.


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