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Countably Dense Subsets in a Metric Space ... Stromberg, Lemma 3.44 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand the proof of Lemma 3.44 on page 105 ... ...


Lemma 3.44 and its proof read as follows:




Stromberg - Lemma 3.44 ... ... .png




In the above proof by Stromberg we read the following:

" ... ... Also, if \(\displaystyle x \in X\) and \(\displaystyle \epsilon \gt 0\), it follows from (2) that \(\displaystyle B_\epsilon (x) \cap A \supset B_{ 1/n } (x) \cap A_{ 1/n } \neq \emptyset\) , where \(\displaystyle 1/n \lt \epsilon\) ... ... "


My question is as follows:

Can someone please demonstrate rigorously why/how it is the case that \(\displaystyle B_\epsilon (x) \cap A \supset B_{ 1/n } (x) \cap A_{ 1/n }\) ... ... ?







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*** EDIT ***

After a little reflection this issue may be straightforward ... ...


Wish to show formally that \(\displaystyle B_{ 1/n } (x) \cap A_{ 1/n } \subset B_\epsilon (x) \cap A \)


We need to show that \(\displaystyle x \in B_{ 1/n } (x) \cap A_{ 1/n } \Longrightarrow x \in B_\epsilon (x) \cap A\)


But ... leaving out details ... we have ...


\(\displaystyle x \in B_{ 1/n } (x) \cap A_{ 1/n }\)


\(\displaystyle \Longrightarrow x \in B_{ 1/n } (x) \text{ and } x \in A_{ 1/n }\)


\(\displaystyle \Longrightarrow x \in B_{ \epsilon } (x) \text{ and } x \in A\)


\(\displaystyle \Longrightarrow x \in B_\epsilon (x) \cap A\)



Is that correct?

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Help will be appreciated ...

Peter
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, it is straight forward. In a metric space, where we have a notion of "d(x,y)", the distance between points x and y, a "ball", "$B_\epsilon(x)$" where x is a point of the metric space and $\epsilon$ is a positive real number, is defined as the set of all points, y, in the metric space such that $d(x,y)< \epsilon$. Geometrically it is the interior of the sphere with center at x and radius $\epsilon$.

Geometrically, $B_x(1/n)$ is the ball with center x and radius 1/n while $B_x(\epsilon)$ is the ball with center x and radius $\epsilon$. If $1/n< \epsilon$ then the two balls have the same center but $B_x(1/n)$ has the smaller radius so is completely contained in $B_x(\epsilon)$.

Algebraically, any point, y, ih $B_x(1/n)$ has distance from x, d(x,y)< 1/n. But $1/n< \epsilon$ so $d(x,y)< 1/n< \epsilon$. Since $d(x,y)< \epsilon$, y is also in $B_x(\epsilon)$. Since y could be any point of $B_x(1/n)$, $B_x(1/n)\subset B_x(\epsilon)$.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
By the way, in your title, "Countably Dense Subsets in a Metric Space", "countably" is an adverb modifying "dense". That makes no sense- there is no such thing as "countably dense". You should have "Countable Dense Subsets in a Metric Space" where "countable" is an adjective modifying "subset".