Proving B(r,x) is a Subset of S^c: Basic Topology Question

[trap101]

Assume $|X| > \rho$ , let $r = |X| - \rho$

Now I am trying to show that $B(r,x)\subseteq S^c$

This should be a simple question, but I am struggling trying to find the right inequlity.

Attempt:

let $y$ be a point in $B(r,x)$.

I know that $|x - y| < r$.

I have to somehow show that $|y| > \rho$

this is where my argument falls apart:

$|y| \leq |y-x| + |x|< r + \rho$ (by triangle inequality)

but this doesn't show that $|y| > \rho$

what am I missing?

 

[HallsofIvy]

It's awfully hard to figure out what you are trying to say here.

Assume |X| > \rho , let r = |X| - \rho

Is "X" a number and |X| its absolute value (as opposed to a vector and its length)? If a vector or point, in R2, R³, or Rⁿ?

Now I am trying to show that B(r, x) \subset S^c

Okay, Is "x" the same as "X", above? I presume that B(r, x) is the ball of radius r centered on x (in some metric space) but what is S^c? I would guess "the compliment of set S" but what is S?

This should be a simple question, but I am struggling trying to find the right inequlity.

Attempt:

let y be a point in B(r,x).

I know that |x - y| < r.

I have to somehow show that |y| > \rho

this is where my argument falls apart:

|y| <= |y-x| + |x| (by triangle inequality) < r + \rho

but this doesn't show that |y| > \rho

what am I missing?

 

[trap101]

It's awfully hard to figure out what you are trying to say here.


Is "X" a number and |X| its absolute value (as opposed to a vector and its length)? If a vector or point, in R<sup>2, R3, or Rn?


Okay, Is "x" the same as "X", above? I presume that B(r, x) is the ball of radius r centered on x (in some metric space) but what is Sc? I would guess "the compliment of set S" but what is S?</sup>

^{I apologize my writing of the question is very messy. You are right on both counts. X is a vector in Rn, Sc is the complement of S, and S is a subset of Rn. Sorry for making it messy.}

 

[HallsofIvy]

Then your question makes no sense. You are given that |x|&gt; \rho, and you want to prove that B(r, x)\subset S^c, the complement of set S? But what is "S". You don't mention it in the hypotheses.

 

[trap101]

Then your question makes no sense. You are given that |x|&gt; \rho, and you want to prove that B(r, x)\subset S^c, the complement of set S? But what is "S". You don't mention it in the hypotheses.

Sorry. S is B(\rho, 0) , that is the ball of radius \rho about the origin.

 

[Fredrik]

OK, you want to show that if $0<\rho<\|x\|$ and $r=\|x\|-\rho$, then $B(x,r)\subseteq B(0,\rho)^c$. Let $y\in B(x,r)$ be arbitrary. You want to show that $\|y\|>\rho$. I suggest this as the first step:
$$\|y\|\geq\|x\|-\|y-x\|.$$

 

[trap101]

OK, you want to show that if $0<\rho<\|x\|$ and $r=\|x\|-\rho$, then $B(x,r)\subseteq B(0,\rho)^c$. Let $y\in B(x,r)$ be arbitrary. You want to show that $\|y\|>\rho$. I suggest this as the first step:
$$\|y\|\geq\|x\|-\|y-x\|.$$

Ok using: $$\|y\|\geq\|x\|-\|y-x\|.$$

I was trying this before but didn't feel it would be valid. Now since $r=\|x\|-\rho$, I can then say

$r + \rho = \|x\|$

now can I say that:

$\|y\|\geq\|x\|-\|y-x\|\geq r + \rho = \|x\| - r$

which wold reduce to $\|y\|\geq\rho$

 

[Fredrik]

Ok using: $$\|y\|\geq\|x\|-\|y-x\|.$$

I was trying this before but didn't feel it would be valid.

Are you familiar with the triangle inequality in the form $\|x+y\|\geq \|x\|-\|y\|$? (This can be derived from the usual version). Write $\|y\|=\|(y-x)+x\|$ and then use this.

$r + \rho = \|x\|$

now can I say that:

$\|y\|\geq\|x\|-\|y-x\|\geq r + \rho = \|x\| - r$

This should be $\|y\|\geq\|x\|-\|y-x\| > r+\rho-r$. (Because $\|x\|=r+\rho$ and $\|y-x\|<r$).

 

[trap101]

This should be $\|y\|\geq\|x\|-\|y-x\| > r+\rho-r$. (Because $\|x\|=r+\rho$ and $\|y-x\|<r$).

Yea I just copied the code wrong for that, but no i was not aware of the triangle inequality in that form. I'm gping to go derive it now. Thanks. If I have an issue I will ask for assistence