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

[Math Amateur]

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:

View attachment 9163In the above text (in the statement of Proposition 6.8 ...) we read the following:

" ... ... $$x \in \overline{E}$$ if and only if $$U \cap E \neq \emptyset$$ for every open neighborhood $$U$$ of $$x$$ (and hence for every neighborhood $$U$$ of $$x$$) ... ..."My question is as follows:

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

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

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

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:
View attachment 9164

Hope that helps ...

Peter

 

[HallsofIvy]

Your question is "why does $x\in \overline E$ and U an open imply $U\cap E$ non-empty"

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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]

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

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

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

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 $W$.

 

[Math Amateur]

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