Prove that a set with an uncountable subset is itself uncountable.

Printable View

- Feb 1st 2009, 02:08 PMninano1205[SOLVED] uncountable subset is itself uncountable
Prove that a set with an uncountable subset is itself uncountable.

- Feb 1st 2009, 02:47 PMPlato
The proof of this really depends upon what you have already proven and how you proved it.

What definitions and theorems you have to work with. That we don’t know.

In most developments of this material, one would normally have already have proven that*Any subset of a countable set is a countable set*. Having done that what is the point of this problem?

If $\displaystyle A \subseteq B$ and $\displaystyle A$ is uncountable there can be no injection $\displaystyle f:B \mapsto \mathbb{N}$.

For if there were its restriction to $\displaystyle A$ would be an injection.

I hope you can see how different many different approaches there are and see that the one you need to use depends on what you have already done. - Feb 3rd 2009, 09:30 AMninano1205Is this statement correct?
From the theorem "Every subset of a countable set is countable", can also mean "a set which has any uncountable subset is itself uncountable."

Is it right? - Feb 3rd 2009, 10:30 AMlandon217
Yes those statements are equivalent. But the very fact that you ask tells you that the connection isn't clear as crystal. And you don't want to trivialize the question if it is a homework problem.

If you want to really show the two statements follow from each other, I suggest write the first in If/then form so you can take the contrapositive. Onces that is done I think you will find your way.