# Thread: metric spaces and infimums

1. ## metric spaces and infimums

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?

2. 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.

3. 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?

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.

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

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?

7. 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?

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?

9. 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?

10. 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?

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