0

$\textbf{Definition}$ : Let $(X,d)$ a metric space, $a\in X$ and $A,B\subseteq X$.

The distance from $x$ to set $A$ is the set : $$ d(x,A)= inf\{ d(x,a);a \in A\} $$ And the distance from $A$ to $B$ is the set : $$ d(A,B)=inf\{d(a,b) \vert a\in A,b\in B \}$$

Let $X$ a metric space and $A,B \subseteq X$. Prove that for $x \in X$ :

$$ d(A,B) \leq d(x,A)+d(x,B) $$

Mi proof :

Let suppose $d(A,B)> d(x,A)+d(x,B) $ then $d(A,B)-d(x,A)>d(x,B)$ so by the definition of $d(x,B)$ exists $b\in B$ such that

$$ d(x,b)<d(A,B)-d(x,A) \implies d(x,A)<d(A,B)-d(x,b) $$

Again by the definition of $d(x,A)$ exists $a\in A$ such that

$$ d(x,a)< d(A,B)-d(x,b) \implies d(a,b) \leq d(a,x)+d(x,b) < d(A,B) $$

A contradiction so $$ d(A,B) \leq d(x,A)+d(x,B) $$

Mi proof is right?

Community
  • 1
  • Please, define what you mean by $d(x,A)$, $d(A,B)$, since there are different notions. It appears that you are using $d(A,B)=inf d(a,b)$ where $inf$ is taken over all pairs $(a,b)\in A\itimes B$. – Moishe Kohan Apr 03 '19 at 17:06
  • $d(x,A)= inf{ d(x,a), a \in A }$ and $d(A,B)= inf { d(a,b); a \in A, b \in B } $ –  Apr 03 '19 at 17:11
  • You should edit your question by adding these definitions. – Moishe Kohan Apr 03 '19 at 17:40

1 Answers1

0

Yes, your proof is correct.

[30 characters]

Klaus
  • 10,578