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The Definition of a Neighborhood and the Definition of an Open Set ... Carothers, Chapters 3 & 4 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
The Definition of a Neighborhood and the Definition of an Open Set ... Carothers, Chapters 3 & 4 ...

I am reading N. L. Carothers' book: "Real Analysis". ... ...

I am focused on Chapter 3: Metrics and Norms and Chapter 4: Open Sets and Closed Sets ... ...

I need help with an aspect of Carothers' definitions of open balls, neighborhoods and open sets ...


Now ... on page 45 Carothers defines an open ball as follows:



Carothers - Defn of an Open Ball ... Page 45 ... .png


Then ... on page 46 Carothers defines a neighborhood as follows:



Carothers - Defn of a Neighborhood of x ... ... page 46 ... .png



And then ... on page 51 Carothers defines an open set as follows:



Carothers - Defn of an Open Set ... Page 51 ... .png



Now my question is as follows:

When Carothers re-words his definition of an open set he says the following:

" ... ... In other words, \(\displaystyle U\) is an open set if, given \(\displaystyle x \in U\), there is some \(\displaystyle \epsilon \gt 0\) such that \(\displaystyle B_\epsilon (x) \subset U \) ... ... "



... BUT ... in order to stay exactly true to his definition of neighborhood shouldn't Carothers write something like ...


" ... ... In other words, \(\displaystyle U\) is an open set if, for each \(\displaystyle x \in U\), \(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\) ... ..."


Can someone lease explain how, given his definition of neighborhood he arrives at the statement ...

" ... ... In other words, \(\displaystyle U\) is an open set if, given \(\displaystyle x \in U\), there is some \(\displaystyle \epsilon \gt 0\) such that \(\displaystyle B_\epsilon (x) \subset U\) ... ... "




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Reflection ... maybe we can regard \(\displaystyle B_\epsilon (x)\) as a neighborhood contained in U since \(\displaystyle B_{ \frac{ \epsilon }{ 2} }(x)\) \(\displaystyle \subset\) \(\displaystyle B_\epsilon (x)\) ... is that correct?


But then why doesn't Carothers just define a neighborhood of \(\displaystyle x\) as an open ball about \(\displaystyle x\) ... rather than a set containing an open ball about \(\displaystyle x\)?


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




Hope someone can clarify ...

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
Re: The Definition of a Neighborhood and the Definition of an Open Set ... Carothers, Chapters 3 & 4

... BUT ... in order to stay exactly true to his definition of neighborhood shouldn't Carothers write something like ...


" ... ... In other words, \(\displaystyle U\) is an open set if, for each \(\displaystyle x \in U\), \(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\) ... ..."
If \(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\) then $x\in B_\epsilon (x) \subseteq N \subseteq U$, so it is certainly true that $B_\epsilon (x) \subseteq U$. Conversely, $B_\epsilon (x)$ is a neighbourhood of $x$, so if $B_\epsilon (x) \subseteq U$ then we can take $N = B_\epsilon (x)$. It will then be true that "\(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\)".

Notice that an open set containing $x$ is a neighbourhood of $x$. But a neighbourhood of $x$ need not be an open set contining $x$ (because a neighbourhood does not have to be open).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Re: The Definition of a Neighborhood and the Definition of an Open Set ... Carothers, Chapters 3 & 4

If \(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\) then $x\in B_\epsilon (x) \subseteq N \subseteq U$, so it is certainly true that $B_\epsilon (x) \subseteq U$. Conversely, $B_\epsilon (x)$ is a neighbourhood of $x$, so if $B_\epsilon (x) \subseteq U$ then we can take $N = B_\epsilon (x)$. It will then be true that "\(\displaystyle U\) contains a neighborhood \(\displaystyle N\) of \(\displaystyle x\) such that \(\displaystyle N\) contains an open ball \(\displaystyle B_\epsilon (x)\)".

Notice that an open set containing $x$ is a neighbourhood of $x$. But a neighbourhood of $x$ need not be an open set contining $x$ (because a neighbourhood does not have to be open).


Thanks for the help Opalg ...

Peter