MHB Proving d(x,A) ≤ d(x,y) + d(y,A) in Metric Spaces

[ozkan12]

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)d(x,A)=infz∈Ad(x,z). Now, say z0∈A and y∈X. Then d(x,z0)≤d(x,y)+d(y,z0). Taking infimum over all z∈A of the left hand side, we obtain:

d(x,A)=infz∈Ad(x,z)≤d(x,z0)≤d(x,y)+d(y,z0).

Observe that d(x,A) is now independent of z0. Hence taking the infimum over all z in A of the right hand side, we get:

d(x,A)≤d(x,y)+infz∈Ad(y,z)=d(x,y)+d(y,A).

İn this proof, we take inf for d(y,z0). but why we didnt take inf for d(x,y) ?

 

[Evgeny.Makarov]

İn this proof, we take inf for d(y,z0). but why we didnt take inf for d(x,y) ?

Because infimum is taken over $z_0$, and $d(x,y)$ does not depend on $z_0$.

The proof uses three properties of infimum.

1. If $u\in B$, then $\inf B\le u$.
2. If $v\le u$ for all $u\in B$, then $v\le\inf B$.
3. $\inf_{u\in B}(c+u)=c+\inf B$ where $c$ does not depend on $u$.

(You may observe that properties 1 and 2 form the definition of infimum. Property 3 is proved using the interaction between $+$ and $\le$.)

Starting from $d(x,z_0)\le d(x,y)+d(y,z_0)$ and using property 1 where $u=d(x,z_0)$ and $B=\{d(x,z)\mid z\in A\}$, we get
\[
d(x,A)\overset{\text{def}}{=}\inf_{z\in A} d(x,z)\le d(x,z_0)\le d(x,y)+d(y,z_0).
\]
This inequality holds for all $z_0\in A$. Now applying property 2 where $v=d(x,A)$ and $B=\{d(x,y)+d(y,z_0)\mid z_0\in A\}$, we get
\[
d(x,A)\le\inf_{z\in A} (d(x,y)+d(y,z)).
\]
Finally, applying property 3 we get
\[
d(x,A)\le d(x,y)+\inf_{z\in A} d(y,z)\overset{\text{def}}{=}d(x,y)+d(y,A).
\]

P.S. I suggest you click the "Reply With Quote" button and study the LaTeX formatting in my post. It's not difficult, and you will be able to type your questions using beautiful math.