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$$\text{d}_{H}(A,B) = \max\left\{ \sup_{a\in A} \inf_{b\in B} \text{d}(a,b),\sup_{b\in B} \inf_{a\in A}\text{d}(a,b)\right\}$$

where $A$ and $B$ are two closed subsets of a metric space $(E,d)$ is a pseudo-distance if $\text{d}_{h}(A,B)=0 \implies$ the RHS of the above equation is 0 As the maximum of two positive elements is always zero($d(a,b) \geq 0$) Both the terms inside the max are zero ie $\sup_{a \in A} \inf_{b \in B} \text{d}_{h}(a,b)=0$ But as the supremum over the elements of $A$ is zero $\implies$

$\inf_{b \in B} \text{d}_{h}(a,b)=0 \implies d(B,a)=0$, where $d(B,a)=0$ is the distance of a point of $a$ from set B(It is defined in such a way) and it follows that $a$ is a contact point of $B$. But this is true $\forall a \in A$, we have $\forall a \in A$, we have $ a \in \overline{B}$ where $\overline{B}$ is the set of all contact points of B which is a closed set $ \implies A\subset \overline{B}$

Similarly we can show that $$B\subset \overline{A}$$

Now the above two imply that $\overline {A}=\overline{B}$

Clearly I could not prove A=B , following from above line of reasoning. Could someone show me why A is not equal to B? I can't think of a counter example where $\overline {A}=\overline{B}$ does not imply $A=B$

Andrés E. Caicedo
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user3503589
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1 Answers1

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If you assume that $A$ and $B$ are closed, as you indicate, then $A = \bar A$ and $B = \bar B$. Thus $A = B$.

If you drop the assumption that $A$ and $B$ are closed, then $d_H$ is just a pseudo-metric. For example, $\mathbb{Q}$ and $\mathbb{Q} + \sqrt{2}$ are different and at distance zero. Of course they have the same closure.

Giovanni Mascellani
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