# Neighbourhoods and Open Neighbourhoods ... Browder, Proposition 6.8 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am reading Chapter 6: Topology ... ... and am currently focused on Section 6.1 Topological Spaces ...

I need some help in order to fully understand an aspect of Proposition 6.8 ... ...

Proposition 6.8 (and the relevant Definition 6.8 ... ) read as follows: In the above text (in the statement of Proposition 6.8 ...) we read the following:

" ... ... $$\displaystyle x \in \overline{E}$$ if and only if $$\displaystyle U \cap E \neq \emptyset$$ for every open neighborhood $$\displaystyle U$$ of $$\displaystyle x$$ (and hence for every neighborhood $$\displaystyle U$$ of $$\displaystyle x$$) ... ..."

My question is as follows:

Why, if the statement: " ... $$\displaystyle x \in \overline{E}$$ if and only if $$\displaystyle U \cap E \neq \emptyset$$ .. "

... is true for every open neighborhood $$\displaystyle U$$ of $$\displaystyle x$$ ...

... is the statement necessarily true for every neighborhood $$\displaystyle U$$ of $$\displaystyle x$$ ... ?

Help will be appreciated ...

Peter

=====================================================================================

The definition of a neighborhood is relevant to the above post ... so I am providing access to Browder's definition of the same as follows: Hope that helps ...

Peter

#### HallsofIvy

##### Well-known member
MHB Math Helper
Your question is "why does $x\in \overline E$ and U an open imply $U\cap E$ non-empty"

Okay, $\overline{E}$ is the "closure of E", the intersection of all closed sets that contain E. And a set is "closed" if it contains all of its boundary points with a "boundary point" being a point, p, such that every open set containing p also contains at least one point of E.

#### Opalg

##### MHB Oldtimer
Staff member
Why, if the statement: " ... $$\displaystyle x \in \overline{E}$$ if and only if $$\displaystyle U \cap E \neq \emptyset$$ .. "

... is true for every open neighborhood $$\displaystyle U$$ of $$\displaystyle x$$ ...

... is the statement necessarily true for every neighborhood $$\displaystyle U$$ of $$\displaystyle x$$ ... ?
If $W$ is a neighbourhood of $x$ then (by definition) it contains an open neighbourhood of $x$. That open neighbourhood has a nonempty intersection with $E$, hence so does $W$.

#### Peter

##### Well-known member
MHB Site Helper
If $W$ is a neighbourhood of $x$ then (by definition) it contains an open neighbourhood of $x$. That open neighbourhood has a nonempty intersection with $E$, hence so does $W$.

HallsofIvy and Opalg ... thanks for the help

Peter