Hausdorff distance and compact sets.

[woven laurel]

I have an answer for the first part of the question, thank you for verifying if the answer is right, and also help me with the example

English traduction :
Let ( K_1 ) and ( K_2 ) be two disjoint compact subsets of a metric space ( (M, d) ). Show that the Hausdorff distance between ( K_1 ) and ( K_2 ), defined by
[
d_H(K_1, K_2) = \inf {d(x_1, x_2) \mid x_1 \in K_1, x_2 \in K_2},
]
is always strictly positive. Find an example of two disjoint non-compact subsets ( K_1 ) and ( K_2 ) such that ( d_H(K_1, K_2) = 0 ).

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[patent tulip]

Vermithor

[woven laurel]

That's the answer I have found :

To prove that ( d_H(K_1, K_2) > 0 ), observe the following:

• Since ( K_1 ) and ( K_2 ) are disjoint, ( d(x_1, x_2) > 0 ) for all ( x_1 \in K_1 ) and ( x_2 \in K_2 ).
• Furthermore, compactness ensures that the infimum is achieved as a minimum because the function ( d(x_1, x_2) ) is continuous and the product ( K_1 \times K_2 ) is compact.
• Therefore, there exists ( (x_1, x_2) \in K_1 \times K_2 ) such that ( d_H(K_1, K_2) = d(x_1, x_2) > 0 ).

Thus, ( d_H(K_1, K_2) > 0 ) when ( K_1 ) and ( K_2 ) are compact and disjoint.

[patent tulip]

Vermithor

[upbeat badge]

Yep seems good to me

Compactness and continuity of (x,y)—>d(x,y) on K1 x K2, ensures that the inf is attained

[woven laurel]

perfect, thank you so much, what about an exemple of two disjoint non compact subsets K1 and K2 such that d(K1,K2) = 0 ? what should I think of ?

[upbeat badge]

Think a dense subset and it’s complementary for example

like in R, Q and R\Q

[woven laurel]

You have always the best exemples ever,
thanks a lot !

Considérons l'exemple suivant :

( K_1 = {x \in \mathbb{R} \mid x > 0}, \quad K_2 = {x \in \mathbb{R} \mid x < 0}. )

Ces deux sous-ensembles sont :

• Non compacts : ils ne sont pas bornés ni fermés.
• Disjoints : ( K_1 \cap K_2 = \emptyset ).

Calculons la distance de Hausdorff ( d_H(K_1, K_2) ). Par définition :

[
d_H(K_1, K_2) = \inf { d(x_1, x_2) \mid x_1 \in K_1, , x_2 \in K_2 }.
]

Pour tout ( \epsilon > 0 ), il existe ( x_1 \in K_1 ) et ( x_2 \in K_2 ) tels que :

[
|x_1 - x_2| < \epsilon.
]

Par exemple, en choisissant ( x_1 = \frac{\epsilon}{2} ) et ( x_2 = -\frac{\epsilon}{2} ), nous avons :

[
|x_1 - x_2| = \left| \frac{\epsilon}{2} - \left(-\frac{\epsilon}{2}\right) \right| = \epsilon.
]

Cela montre que ( d_H(K_1, K_2) = 0 ).

[patent tulip]

Vermithor

[woven laurel]

Chatgpt proposed this as well, seems so right to me, what do u think ?

[upbeat badge]

Yeah works as well

[woven laurel]

+close

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