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

#### Peter

##### Well-known member
MHB Site Helper
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:

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
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
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".