# Cardinals and order problems

• Apr 9th 2010, 07:05 AM
Ester
Cardinals and order problems
I have problems with cardinals and their order.
Problem:
Let K,L,M be sets so that card(K) = k, card(L) = l and card(M) = m.
Show that from the condition k < l does not follow, that
a) k + m < l + m
b) k*m < l*m
c) m^k < m^l

(Notice: the symbol is really "smaller than" not "smaller or as great as".)

I have tried to solve this for hours, and I really don't get a grip of it. Can anyone even help me to right direction?
• Apr 9th 2010, 07:28 AM
MoeBlee
a) k + m < l + m

Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.

b) k*m < l*m

Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.

c) m^k < m^l

Hint: Let m = 0. Then find suitable k and l to contradict the above.

Or another approach:

Hint: You're allowed to let any of those cardinals be finite or infinite in order to contradict the above.
• Apr 9th 2010, 10:01 PM
Ester
Okey, it didn't even come to my mind, that I can assume some of the cardinals to be infinite.

I was thinking to proceed like this: "Let k be finite and l be infinite...", but I still don't manage to solve the problems.
• Apr 10th 2010, 07:34 AM
MoeBlee
Quote:

Originally Posted by Ester
Let k be finite and l be infinite

Not quite. Instead, think about this question (your course or book should have covered this by the time you got to this excercise): What happens when you add a finite cardinal to an infinite cardinal?
• Apr 12th 2010, 04:48 AM
Ester
Quote:

Originally Posted by MoeBlee
What happens when you add a finite cardinal to an infinite cardinal?

Okey.. Let k be finite cardinal and l infinite cardinal.
I think that happens something like: k + l = l. Am I correct?
• Apr 12th 2010, 06:51 AM
MoeBlee
Quote:

Originally Posted by Ester
Okey.. Let k be finite cardinal and l infinite cardinal.
I think that happens something like: k + l = l. Am I correct?

Yes, in the sense that a finite cardinal added to an infinite cardinal is just the infinite cardinal. But in this case, it's not l that you want to be infinite. Now just determine which of k, l, and m should be finite and infinite and you're almost done.
• Apr 12th 2010, 07:37 AM
Ester
Quote:

Originally Posted by MoeBlee
Now just determine which of k, l, and m should be finite and infinite and you're almost done.

I have this condition: k < l, so in my mind, if I assume that l is finite cardinal, then k also have to be finite, because k < l < N (aleph). So the m should be infinite..?
• Apr 12th 2010, 07:48 AM
MoeBlee
Right. Let k < l both finite and m infinite.
• Apr 12th 2010, 08:03 AM
Ester
Is it really this easy: Let k,l be finite, and m infinite. Now k + m = m = l + m.
And so it can't be k + m < l + m.

Now I think I have to prove that there is bijection between k + m and m, because we don't have this kind of theory (finite + infinite = infinite) in our course.
• Apr 12th 2010, 08:13 AM
MoeBlee
Yes, if you also say that you've chosen k < l.

So I would write it this way:

Let k and l both be finite with k < l. Let m be infinite. Then k+m = l+m, so it is not the case that k+m < l+m. So it is not the case that, for all cardinals k, l, and m, we have k < l implies k+m < l+m.

But remember, this depends on having already proven (in your course or in your textbook) that a finite k added to an infinite m is just m.