# Basic topology question

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?

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
Homework Helper
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) $\subset$ Sc
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?

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?

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 R2, 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?

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
Homework Helper
Then your question makes no sense. You are given that $|x|> \rho$, 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.

Then your question makes no sense. You are given that $|x|> \rho$, 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
Staff Emeritus
Gold Member
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, 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
Staff Emeritus
Gold Member
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##).

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