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

Peter

Well-known member
MHB Site Helper
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:

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

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

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 $$\displaystyle x$$?

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Hope someone can clarify ...

Peter

Last edited:

Opalg

MHB Oldtimer
Staff member
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
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