Carothers' Definitions: Neighborhoods, Open Sets, and Open Balls

[Math Amateur]

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:
View attachment 9213Then ... on page 46 Carothers defines a neighborhood as follows:
View attachment 9214
And then ... on page 51 Carothers defines an open set as follows:
View attachment 9215
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

 

[Opalg]

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).

 

[Math Amateur]

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