metric spaces and infimums

**choccookies** · November 11th 2010, 06:42 PM

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?

**Tinyboss** · November 11th 2010, 06:53 PM

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.

**choccookies** · November 13th 2010, 08:21 AM

could i do this:

inf{dist(x,J): x member of Q}
= inf{inf{d(x,y)|y member of J}:x member of Q}

Since we know inf{d(x,y)|x member of Q, y member of J}= d(Q,J)

Then the above is = inf{d,(Q,J)},

and then the same for the right hand side?

**Drexel28** · November 13th 2010, 11:57 AM

Originally Posted by choccookies

could i do this:

inf{dist(x,J): x member of Q}
= inf{inf{d(x,y)|y member of J}:x member of Q}

Since we know inf{d(x,y)|x member of Q, y member of J}= d(Q,J)

Then the above is = inf{d,(Q,J)},

and then the same for the right hand side?

$\inf \{d(Q,J)\}$ makes very little sense.

Try proving that each infimum is a lower bound of the other set.

**choccookies** · November 13th 2010, 12:55 PM

So I shouldnt use any of what i wrote?
Could you start me off on how to prove that?

**Drexel28** · November 13th 2010, 01:07 PM

Originally Posted by choccookies

So I shouldnt use any of what i wrote?
Could you start me off on how to prove that?

So, if $\alpha=\inf\left\{d(x,J):x\in Q\right\}$ and $\beta=\inf \left\{d(Q,y):y\in J\right\}$ then you're really trying to prove that $\alpha\leqslant \beta$ and $\beta\leqslant \alpha$ . But, by definition it suffices to show that $\alpha\leqslant \inf\left\{d(Q,y):y\in J\right\}$ and $\beta\leqslant\left\{d(x,J):x\in Q\right\}$ . Can you work with that?

**choccookies** · November 13th 2010, 01:50 PM

Originally Posted by Drexel28

So, if $\alpha=\inf\left\{d(x,J):x\in Q\right\}$ and $\beta=\inf \left\{d(Q,y):y\in J\right\}$ then you're really trying to prove that $\alpha\leqslant \beta$ and $\beta\leqslant \alpha$ . But, by definition it suffices to show that $\alpha\leqslant \inf\left\{d(Q,y):y\in J\right\}$ and $\beta\leqslant\left\{d(x,J):x\in Q\right\}$ . Can you work with that?

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?

**Drexel28** · November 13th 2010, 02:01 PM

Originally Posted by choccookies

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!

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).

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?

We can't. That's exactly what I am saying. Recall that if $\varnothing\subsetneq E\subseteq\mathbb{R}$ is bounded from below then $x\leqslant \inf E$ if and only if $x$ is a lower bound for $E$ .

So, to show that $\alpha\leqslant \overbrace{\inf\left\{d(Q,y):y\in J\right\}}^{\beta}$ it suffices to show that $\alpha$ is a lower bound for $\left\{d(Q,y):y\in J\right\}$ and similarly to show that $\beta\leqslant\overbrace{\inf\left\{d(x,J):x\in Q\right\}}^{\alpha}$ it suffices to show that $\beta$ is a lower bound for $\left\{d(x,J):x\in Q\right\}$ .

But, to make things easier (we're back to the general case now) recall that to show that $x$ is a lower bound of $E$ it suffices to show that $x\leqslant e$ for all $e\in E$ .

Can you apply this?

**Plato** · November 13th 2010, 02:28 PM

Let’s simplify the notation.
Let $\alpha=\inf\{D(J,x):x\in Q\}~\&~\beta=\{D(Q,y):y\in J\}$ .
Suppose that $\alpha <\beta$ . So $\left( {\exists q' \in Q} \right)\left[ {\alpha \leqslant D(J,q') < \beta }\right]$ .

Suppose that $\beta <\alpha$ . So $\left( {\exists j' \in J} \right)\left[ {\beta \leqslant D(Q,j') < \alpha }\right]$ .

But note that means $\beta\leqslant d(j',q')~\&~\alpha\leqslant d(j',q')$ .

Do you see a contradiction there?

**choccookies** · November 13th 2010, 05:05 PM

Originally Posted by Drexel28

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 $\varnothing\subsetneq E\subseteq\mathbb{R}$ is bounded from below then $x\leqslant \inf E$ if and only if $x$ is a lower bound for $E$ .

So, to show that $\alpha\leqslant \overbrace{\inf\left\{d(Q,y):y\in J\right\}}^{\beta}$ it suffices to show that $\alpha$ is a lower bound for $\left\{d(Q,y):y\in J\right\}$ and similarly to show that $\beta\leqslant\overbrace{\inf\left\{d(x,J):x\in Q\right\}}^{\alpha}$ it suffices to show that $\beta$ is a lower bound for $\left\{d(x,J):x\in Q\right\}$ .

But, to make things easier (we're back to the general case now) recall that to show that $x$ is a lower bound of $E$ it suffices to show that $x\leqslant e$ for all $e\in E$ .

Can you apply this?

so i need to show that alpha is less than or equal to all elements in {d(Q,y):y member of J}?
Also I just wanted to ask, have you just replaced 'dist' with the metric d throughout this question?

**Drexel28** · November 13th 2010, 05:16 PM

Originally Posted by choccookies

so i need to show that alpha is less than or equal to all elements in {d(Q,y):y member of J}?
Also I just wanted to ask, have you just replaced 'dist' with the metric d throughout this question?

Yes and yes in the sense that I call $d(x,J)=\text{dist}(x,J)$

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