pick out the true statement in metric spaces?

Question

Let {X, d} be a metric space and let E ⊂ X. For x ∈ X, define

d(x, E) = inf d(x,y) where y ∈ E

Pick out the true statements:

(a) |d(x, E) − d(y, E)| ≤ d(x, y) for all x and y ∈ X.

(b) d(x, E) = d(x, m) for some m ∈ E

My attempt : i was taking X = R-{O} , E =(0,∞) and x= -1

        now for option (a) by triangle inequality  it will satisfied.....and 
   it is correct

i don't know about the second option (b).

Pliz help me and tell me the solution,,,,,i would be thankful

I don't understand how you are using the Triangle Inequality...you are correct that the statement is true but I think you need to strengthen your argument. — lulu, Oct 01 '17 at 14:32
E may be closed or open,,,,,,,not sure it is closed @ salahamam-Fatima — , Oct 01 '17 at 14:35
i mean d(x,E) - d(y,E) ≤ d(x,y) ≤ d(x,E) + d(y,E) @ lulu — , Oct 01 '17 at 14:38

score 0 · Answer 1 · answered Oct 01 '17 at 14:51

0

You have come to the correct conclusion about statement (a) albeit for the wrong reasons. A valid line of reasoning is as follows:

Given $z \in E$, we note that $$ d(x,z) \leq d(x,y) + d(y,z) $$ Since this is true for all $z \in E$, we can conclude that $$ d(x,E) = \inf_{z \in E} d(x,z) \leq \inf_{z \in E} [d(x,y) + d(y,z)] = d(x,y) + \inf_{z \in E} d(y,z) = d(x,y) + d(y,E) $$ using this "standard triangle inequality", we can deduce the "reverse triangle inequality".

As for part (b): consider $x = 0$ and $E = (1,2)$ under the usual metric for $\Bbb R$.

answered Oct 01 '17 at 14:51

Ben Grossmann

225,327

thanks a lots @ omnomnomnom,,,but for option b) ,,acording to ur answer d(x,E) = d(x,m) = 0...so option B is also correct ...Am i right ? – Oct 01 '17 at 14:56
No, not if E is open. For example take E= (1, 2) in R and x= 0. Then for any p in E, d(x, p)= |p- 0|= p since p is positive so d(x, E)= inf (|p|) = 1. But 1 itself is not in the open interval (1, 2) so for all p in (1, 2), |p|> 1. – user247327 Oct 01 '17 at 15:00
im not getting ,, how can u claim that d(x, E) = inf(P) = 1 @user247327 – Oct 01 '17 at 15:05
@kalomlego No, for my example, $d(x,E) = 1$. Note that there is no $m \in E$ such that $d(x,m) = 1$. – Ben Grossmann Oct 01 '17 at 15:10
@ omnomnomnom..one last doubt suppose i take m = (4/3,3/2) ∈ (1,2)...in this case i think d(x,E) = d(x,m),,,,,for some m ∈ E – Oct 01 '17 at 15:19
$(4/3,3/2)$ is not an element of $(1,2)$; both of these are intervals. In any case, we have $d(x,m) = 4/3 \neq d(x,E)$. In particular, note that $1 \notin E$. – Ben Grossmann Oct 01 '17 at 15:22
okkk,,,thanks a lot once again @ omnomnomnom.. im voting up ur answer now – Oct 01 '17 at 15:26

pick out the true statement in metric spaces?

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