Complete, Equivalent, Closed sets

[Somefantastik]

If a set A and a set B are equivalent, and it is known that A is complete, can it then be said that B is also complete?

What if it is known that A is closed, can it then be said that B is also closed?

 

[SW VandeCarr]

If a set A and a set B are equivalent, and it is known that A is complete, can it then be said that B is also complete?

What if it is known that A is closed, can it then be said that B is also closed?

Two sets are equal if they have the same elements. So the set A=B={[0,1]} is closed because it contains both 0 and 1. I don't know if you mean something different by using the term "equivalent".

 

[Somefantastik]

Kolmogorov states that equivalents sets are those on which a one-to-one correspondence can be found.

 

[SW VandeCarr]

Kolmogorov states that equivalents sets are those on which a one-to-one correspondence can be found.

This means two sets have the same cardinality. Since both the open and closed sets on the real interval 0,1 have the same cardinality, A and B may nevertheless differ in terms of closure. If you wish to call the closed interval "complete" then the open interval would not be "complete".

 

[Somefantastik]

Since both the open and closed sets on the real interval 0,1 have the same cardinality

Can you clarify that statement a little?

Also, would making the requirement more strict, maybe saying if the sets A and B are isometric, then A closed implies B is closed?

 

[CRGreathouse]

Can you clarify that statement a little?

If a < b, then |[a, b]| = |(a, b)| = |(a, b]| = |[a, b)| where |...| is the cardinality.

 

[SW VandeCarr]

Also, would making the requirement more strict, maybe saying if the sets A and B are isometric, then A closed implies B is closed?

What metric are you defining for your sets? The only inherent measure of a set is the number of elements it contains (ie its cardinality). The fact that sets A and B have the same cardinality doesn't imply that if A is closed, B must be closed.

 

[Somefantastik]

Nevermind, I have the answer...it works when you have a metric preserving homeomorphism between the sets.