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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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:




Browder - Defn of Closure 6.7 and Relevant Propn 6.8  ... .png





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:



Browder - 2 - Start of 6.1 - Relevant Defns & Propns ... PART 2 ... .png




Hope that helps ...

Peter
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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
Feb 7, 2012
2,681
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
Jun 22, 2012
2,891
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