Prove that if A is contained in some closed ball, then A is bounded.

  • Thread starter Hodgey8806
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  • #1
Hodgey8806
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Homework Statement


Let M be a metric space and A[itex]\subseteq[/itex]M be any subset:
Prove that if A is contained in some closed ball, then A is bounded.


Homework Equations


Def of closed-ball: [itex]\bar{B}[/itex]R(x) = {y[itex]\in[/itex]M:d(x,y)≤R} for some R>0
Def of bounded: A is bounded if [itex]\exists[/itex]R>0 s.t. d(x,y)≤R [itex]\forall[/itex]x,y[itex]\in[/itex]A
Empty set is defined to be bounded in for these problems

The Attempt at a Solution


Spse that A is contained in some closed ball.
Let that ball be [itex]\bar{B}[/itex]R(y0) = {y[itex]\in[/itex]M:d(y,y0)≤R}, for some arbitrary fixed y0
1) If A=[itex]\phi[/itex], vacuously true.
2)Let A[itex]\subseteq[/itex][itex]\bar{B}[/itex]R(y0)
Let x1,x2[itex]\in[/itex]A.
The d(x1,x2)≤d(x1,y0) + d(x2,y0)≤2R.
Thus, diam(A)≤2R and we see that A is bounded.
Q.E.D.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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So this is homework but you don't know any relevant formulas or facts? And you haven't made any attempt to do the problem yourself? Are you sure you are taking this class!?

What is the definition of "bounded set"? What is the definition of "closed ball"? If you know those, then this should be relatively easy. If you don't, nothing I say will help you until you have looked up those definitions (different texts may have slightly different definitions- you need to know definitions you are to use.
 
  • #3
Hodgey8806
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Sorry, I accidentally pressed "Enter" after the title lol. Please, see my attempt. I'm pretty sure it is correct; but, I want to make sure that I don't have to say anymore or elaborate on the empty set bit.

Thank you!
 
  • #4
christoff
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Spse that A is contained in some closed ball.
Let that ball be [itex]\bar{B}[/itex]R(y0) = {y[itex]\in[/itex]M:d(y,y0)≤R}, for some arbitrary fixed y0
1) If A=[itex]\phi[/itex], vacuously true.
2)Let A[itex]\subseteq[/itex][itex]\bar{B}[/itex]R(y0)
Let x1,x2[itex]\in[/itex]A.
The d(x1,x2)≤d(x1,y0) + d(x2,y0)≤2R.
Thus, diam(A)≤2R and we see that A is bounded.
Q.E.D.

Yes, your proof is correct. I don't think you need to elaborate on the empty set case.
 
  • #5
Hodgey8806
145
3
Thank you very much! I appreciate your help!
 

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