Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

if we take a {1, 2, 3} set as a countable set.
it has subsets with 2 elements right, Those sets are : {1, 2}, {1, 3}, {2, 3}.
But what about uncountable sets?
if we take a (0, 9) set as subset of real numbers or rational numbers which has many elements, how can i get subsets with two elements of this uncountable set?

share|improve this question
    
Firstly, as a subset of the rationals, $(0,9)$ is countable. But to pull out only two elements, I might choose, say, $\{ 1, 2 \}$, as they're certainly in that set. –  mixedmath Oct 28 '12 at 15:58
    
I don't understand why you retagged this like that. It has nothing to do with functional analysis, and it's much more [elementary-set-theory] than [set-theory]. –  Asaf Karagila May 10 '13 at 9:35

1 Answer 1

up vote 3 down vote accepted

Suppose $A$ is an uncountable set. In particular it is infinite. Fix $k-1$ elements from $A$, say $a_1,\ldots,a_{k-1}$.

Now for every $a\in A\setminus\{a_1,\ldots,a_{k-1}\}$ the function $a\mapsto\{a,a_1,\ldots,a_{k-1}\}$ is injective. Again, since $A$ is uncountable we have that $A\setminus\{a_1,\ldots,a_{k-1}\}$ is equipotent with $A$.

Therefore there are at least $|A|$ many subsets of size $k$; on the other hand we have that $|\{B\subseteq A\mid B\text{ is finite}\}|=|A|$ as well. While not a completely trivial theorem, it is still a true theorem for any infinite set. Therefore there cannot be more than $|A|$ sets of size $k$.

Therefore the cardinality of $\{B\subseteq A\mid |B|=k\}$ is exactly $|A|$.

(Note that I am assuming the axiom of choice holds, or at least that $A$ can be well-ordered. This is not an uncommon assumption in modern mathematics. The question, when asked in the broader context where the axiom of choice fails and $A$ cannot be well-ordered is vastly more complicated to answer due to all sort of pathological sets which may arise.)

share|improve this answer
2  
Not an uncommon assumption is definitely a litotes! –  Brian M. Scott Oct 28 '12 at 17:52
    
@Brian: It is not not not not a litotes, you meant to say! :-) (I'm assuming the real life has double negation elimination, by the way...) –  Asaf Karagila May 10 '13 at 10:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.