# Open Balls in a Normed Vector Space .... Carothers, Exercise 32

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In summary: Thanks for the help ... I'm still thinking over the proof, but I think the idea is that for any given $x,y\in V$ there exists a unique $z\in B_r(0)$ such that $y-x=z$. Then $y=x+z$ holds in $V$, since $x+B_r(0) = V$.
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MHB
I am reading N. L. Carothers' book: "Real Analysis". ... ...

I am focused on Chapter 3: Metrics and Norms ... ...

I need help Exercise 32 on page 46 ... ... Exercise 32 reads as follows:

View attachment 9201
I have not been able to make much progress ...

We have ...$$\displaystyle B_r(x) = \{ y \in M \ : \ d(x, y) \lt r \}$$

... and ...

$$\displaystyle B_r(0) = \{ y \in M \ : \ d(0, y) \lt r \}$$... and ...$$\displaystyle x + B_r(0) = x + \{ y \in M \ : \ d(0, y) \lt r \}$$But ... how do we formally proceed from here ...
Hope that someone can help ...

Peter===================================================================================

The above post refers to/involves the notion of an open ball ... so I am providing Carothers' definition of the same ... as follows:
View attachment 9202
The above post also refers to/involves the notion of a normed vector space ... so I am providing Carothers' definition of the same ... as follows:
View attachment 9203
View attachment 9204Hope that helps ...

Peter

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Given $x,y\in V$, let $z = y-x$. Then $$y \in B_r(x) \Longrightarrow \|y-x\|<r \Longrightarrow \|z\|<r \Longrightarrow z\in B_r(0) \Longrightarrow y = x+z \in x + B_r(0).$$ Therefore $B_r(x) \subseteq x + B_r(0)$. That argument also works in the opposite direction, giving the reverse inclusion in the same way.

Opalg said:
Given $x,y\in V$, let $z = y-x$. Then $$y \in B_r(x) \Longrightarrow \|y-x\|<r \Longrightarrow \|z\|<r \Longrightarrow z\in B_r(0) \Longrightarrow y = x+z \in x + B_r(0).$$ Therefore $B_r(x) \subseteq x + B_r(0)$. That argument also works in the opposite direction, giving the reverse inclusion in the same way.
Hi Opalg ...

Thanks for the help ...

... but ... just a clarification ...

You write:

" ... ... $$\displaystyle z\in B_r(0) \Longrightarrow y = x+z \in x + B_r(0)$$ ... ... "I am somewhat perplexed as to how we justify (or even interpret ...) the above statement ... how do you justify it based solely on the axioms of a normed vector space and the basics of set theory ... indeed ... what does the statement ... ... $$\displaystyle y = x+z \in x + B_r(0)$$ ... ... mean exactly ...
My thoughts are as follows ...

$$\displaystyle y = x + z$$

$$\displaystyle \Longrightarrow y = x \ +$$ ... a particular element of the set $$\displaystyle B_r(0)$$

$$\displaystyle \Longrightarrow$$ ... (means ... ) ... $$\displaystyle y \in x + B_r(0)$$Is that correct ... am I interpreting things correctly ...?

Regarding the reverse argument ... my thoughts are as follows:

$$\displaystyle y \in x + B_r(0)$$

$$\displaystyle \Longrightarrow y = x + z$$ where $$\displaystyle z \in B_r(0)$$

$$\displaystyle \Longrightarrow y - x = z$$ where $$\displaystyle z \in B_r(0)$$But $$\displaystyle \| z \| \lt r$$

$$\displaystyle \Longrightarrow \| y - x \| \lt r$$

$$\displaystyle \Longrightarrow y \in B_x(r)$$ Is that correct?Hope you can help further ...

Peter

Last edited:
Peter said:
You write:

" ... ... $$\displaystyle z\in B_r(0) \Longrightarrow y = x+z \in x + B_r(0)$$ ... ... "I am somewhat perplexed as to how we justify (or even interpret ...) the above statement ... how do you justify it based solely on the axioms of a normed vector space and the basics of set theory ... indeed ... what does the statement ... ... $$\displaystyle y = x+z \in x + B_r(0)$$ ... ... mean exactly ...
My thoughts are as follows ...

$$\displaystyle y = x + z$$

$$\displaystyle \Longrightarrow y = x \ +$$ ... a particular element of the set $$\displaystyle B_r(0)$$

$$\displaystyle \Longrightarrow$$ ... (means ... ) ... $$\displaystyle y \in x + B_r(0)$$Is that correct ... am I interpreting things correctly ...?
That is correct. In fact, in the statement of Exercise 32, Carothers defines $x+B_r(0)$ to mean $\{x+y:\|y\|<r\}$:
View attachment 9205

Peter said:
Regarding the reverse argument ... my thoughts are as follows:

$$\displaystyle y \in x + B_r(0)$$

$$\displaystyle \Longrightarrow y = x + z$$ where $$\displaystyle z \in B_r(0)$$

$$\displaystyle \Longrightarrow y - x = z$$ where $$\displaystyle z \in B_r(0)$$But $$\displaystyle \| z \| \lt r$$

$$\displaystyle \Longrightarrow \| y - x \| \lt r$$

$$\displaystyle \Longrightarrow y \in B_x(r)$$ Is that correct?
For the reverse inclusion $x+B_r(0) \subseteq B_r(x)$, you need to start with an element $z\in B_r(0)$. Then $$z\in B_r(0) \Longrightarrow \|z\|<r \Longrightarrow \|x-(x+z)\|<r \Longrightarrow x+z\in B_r(x).$$

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Opalg said:
That is correct. In fact, in the statement of Exercise 32, Carothers defines $x+B_r(0)$ to mean $\{x+y:\|y\|<r\}$:
For the reverse inclusion $x+B_r(0) \subseteq B_r(x)$, you need to start with an element $z\in B_r(0)$. Then $$z\in B_r(0) \Longrightarrow \|z\|<r \Longrightarrow \|x-(x+z)\|<r \Longrightarrow x+z\in B_r(x).$$
Thanks Opalg ...

Still thinking over the reverse inclusion proof ...

Peter

## 1. What is an "Open Ball" in a Normed Vector Space?

An open ball in a normed vector space is a set of all points that are within a certain distance (radius) from a given point. The distance is measured using the norm of the vector space. The open ball is denoted as B(x,r) where x is the center point and r is the radius.

## 2. How is the radius of an open ball determined in a normed vector space?

The radius of an open ball in a normed vector space is determined by the norm of the vector space. The radius is the maximum distance that any point in the open ball can be from the center point. This distance is measured using the norm of the vector space.

## 3. What is the difference between an open ball and a closed ball in a normed vector space?

An open ball includes all points that are within a certain distance (radius) from a given point, while a closed ball includes all points that are exactly at the given distance from the center point. In other words, a closed ball includes the boundary points that are excluded in an open ball.

## 4. How is an open ball used in normed vector spaces?

An open ball is used in normed vector spaces to define a neighborhood around a given point. It is also used in the definition of continuity and convergence of sequences in a normed vector space. Additionally, open balls are used in the proof of the Heine-Borel theorem, which states that a subset of a normed vector space is compact if and only if it is closed and bounded.

## 5. Can an open ball in a normed vector space be empty?

Yes, an open ball in a normed vector space can be empty if the radius is set to 0 or if the center point is not included in the vector space. In this case, there are no points within the given distance from the center point, thus the open ball is empty.

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