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

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The discussion centers on N. L. Carothers' definitions of open sets, neighborhoods, and open balls in his book "Real Analysis." A key point raised is the relationship between these concepts, particularly how an open set can be defined in terms of neighborhoods. It is clarified that if a set U contains a neighborhood N of a point x, which in turn contains an open ball Bε(x), then Bε(x) must also be a subset of U. The distinction is made that while every open set containing x is a neighborhood of x, not all neighborhoods are open sets. This nuanced understanding of the definitions is essential for grasping the foundational concepts in real analysis.
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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, $$U$$ is an open set if, given $$x \in U$$, there is some $$\epsilon \gt 0$$ such that $$B_\epsilon (x) \subset U $$ ... ... "
... BUT ... in order to stay exactly true to his definition of neighborhood shouldn't Carothers write something like ..." ... ... In other words, $$U$$ is an open set if, for each $$x \in U$$, $$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$ ... ..."Can someone lease explain how, given his definition of neighborhood he arrives at the statement ...

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

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

Reflection ... maybe we can regard $$B_\epsilon (x)$$ as a neighborhood contained in U since $$B_{ \frac{ \epsilon }{ 2} }(x)$$ $$\subset$$ $$B_\epsilon (x)$$ ... is that correct?But then why doesn't Carothers just define a neighborhood of $$x$$ as an open ball about $$x$$ ... rather than a set containing an open ball about $$x$$?=========================================================================================

Hope someone can clarify ...

Peter
 

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

Peter said:
... BUT ... in order to stay exactly true to his definition of neighborhood shouldn't Carothers write something like ..." ... ... In other words, $$U$$ is an open set if, for each $$x \in U$$, $$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$ ... ..."
If $$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$ 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 "$$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$".

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

Opalg said:
If $$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$ 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 "$$U$$ contains a neighborhood $$N$$ of $$x$$ such that $$N$$ contains an open ball $$B_\epsilon (x)$$".

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
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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