If d is a metric, then d(x,y)=d(y,x). You can get it from there--suppose the two infima were not equal, and derive a contradiction.
HI Ive got this question I dont know how to do;
Let X be a metric space, and let Q,J be non-empty subsets of X. prove that
inf{dist(x,J):x is a member of Q}= inf{dist(Q,y):y is a member of J}.
I know that the dist(x,J):= inf{d(x,y)|y is a member of J}, but I am not sure how to tackle this question. Any help please?
Thankyou so much for your help. I'm not just trying to get an answer off you!I really am trying to solve it, i just cant get my head around it!
To prove alpha</beta, wouldnt we have to say that beta is also a lower bound of inf{d(Q,y):y member of J}? How can we just assume that?
It's no problem! We all go through problems which just are too much info at first sight (which I can see how this would be one).
We can't. That's exactly what I am saying. Recall that if is bounded from below then if and only if is a lower bound for .To prove alpha</beta, wouldnt we have to say that beta is also a lower bound of inf{d(Q,y):y member of J}? How can we just assume that?
So, to show that it suffices to show that is a lower bound for and similarly to show that it suffices to show that is a lower bound for .
But, to make things easier (we're back to the general case now) recall that to show that is a lower bound of it suffices to show that for all .
Can you apply this?