Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying with no luck to prove:

Let (X,d) be a metric space and A a non-empty subset of X. For x,y in X, prove that

d(x,A) < d(x,y) + d(y,A)

share|improve this question
1  
You won't be able to prove this with strict inequality; for instance, take $A = \{y\}$. You need $\leq$. –  William Feb 12 '12 at 0:49
    
Is this homework? If so, please put the homework tag on it. Try drawing a picture of a set A with two points x and y outside it. –  Stefan Smith Feb 12 '12 at 0:49
    
It will help if you have sitting in front of you the definition of distance from a point to a set. –  Gerry Myerson Feb 12 '12 at 0:53
    
I've drawn all the pictures. But doesn't it come down to some creative trick or using a fact about infimum? I can't see it. –  student's u Feb 12 '12 at 0:53
add comment

1 Answer

If you have $\leq$ instead of $<$, then you have:

$d(x,A) = \inf_{z\in A} d(x,z)$. Now, say $z_0\in A$ and $y\in X$. Then $d(x,z_0)\leq d(x,y) + d(y, z_0)$. Taking infimum over all $z\in A$ of the left hand side, we obtain:

$$ d(x, A) = \inf_{z\in A}d(x,z) \leq d(x,z_0) \leq d(x,y) + d(y, z_0). $$

Observe that $d(x, A)$ is now independent of $z_0$. Hence taking the infimum over all $z$ in $A$ of the right hand side, we get:

$$ d(x,A) \leq d(x,y) + \inf_{z\in A}d(y,z) = d(x,y) + d(y, A). $$

share|improve this answer
    
Isn't there the possibility that d(y,A) < d(y,z0), in which case your last inequality doesn't necessarily hold? –  student's u Feb 12 '12 at 0:59
    
@student'su if $a\le b_\alpha$ for all $\alpha$, then $a\le \inf_\alpha b_\alpha$. –  David Mitra Feb 12 '12 at 1:03
    
I have a constant on the left, which is not smaller than the right hand side for ANY choice of $z_0\in A$. Now try proving rigorously that this is enough to conclude the proof as I did it above (hint: you can always chose a sequence $\{z_n\}_{n\in\mathbb{N}}$ in $A$ such that $d(y,z_n)\rightarrow \inf_{z\in A}d(y,z)$ as $n\rightarrow\infty$). –  William Feb 12 '12 at 1:03
    
@student : The definition itself of the infimum is the following ; the infimum is a lower bound which is greater than or equal to any other lower bound. Since $d(x,A)$ is a lower bound of $\{ d(x,y) + d(y,z_0) \, | \, z_0 \in A \}$, then it is less than or equal to the infimum. In other words, it's a triviality you shouldn't be worrying about. –  Patrick Da Silva Feb 12 '12 at 1:24
    
Thanks all. I understand. I have no idea why this was hard for me. –  student's u Feb 12 '12 at 1:55
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.