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

#### Peter

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

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

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The above post refers to/involves the notion of an open ball ... so I am providing Carothers' definition of the same ... as follows:

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:

Hope that helps ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
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.

#### Peter

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

#### Opalg

##### MHB Oldtimer
Staff member
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 $\{x+y:\|y\|<r\}$:

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 $$z\in B_r(0) \Longrightarrow \|z\|<r \Longrightarrow \|x-(x+z)\|<r \Longrightarrow x+z\in B_r(x).$$

#### Peter

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