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Let $X=\{[a,b]\ a,b \in R$ and $a<b \}$ and Let $y=\{(a,b)\ a,b \in R $ and $ a<b \}$ then we define a metric on X as:

$$d( [a,b], [c,d] )= \inf \{\epsilon >0 : [a,b]\subseteq [c-\epsilon, d+\epsilon] \ \text{and} \ [c,d]\subseteq [a-\epsilon,b+\epsilon]\}.$$

Does $(X,d) $ define a metric space?

Now if we consider the canonical extension of $d$ call it $D$ on $Z:=X\bigcup Y$ is $(Z,D)$ a metric space?

Seirios
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Whats My Name
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  • What have you done so far? Hint (2.) consider $$D([a_0,b_0] , (a_0,b_0))$$ What does this tell you about $D$? – AlexR Sep 23 '13 at 15:54
  • I have proved that the first one is a metric space since it is basically a minimum of a set considering the cases when the 2 intervals overlap or not and for the second one D([a,b],(a,b)) would be 0. Is this correct? – Whats My Name Sep 23 '13 at 15:57
  • Yes, that's correct and what does it say about $D$ if $x\neq y$ but $D(x,y) = 0$? – AlexR Sep 23 '13 at 16:13
  • It contradicts the distance definition so it is not a metric space on Z. – Whats My Name Sep 23 '13 at 16:17
  • And voila you're done with both tasks ;-) – AlexR Sep 23 '13 at 17:08

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