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Taking finite powers of a countably infinite set still yields a countable set: $$|\mathbb{N}|=|\mathbb{N}^k|.$$

It's also known that countable powers of the continuum still have the same cardinality as the continuum: $$|2^\mathbb{N}|=|\mathbb{R}|=|\mathbb{R}^k|=|\mathbb{R}^\mathbb{N}|.$$

Furthermore, taking the next highest ordinal to the power of the continuum doesn't seem to change it. (at least according to wikipedia) Denote $X:=2^\mathbb{R}$, then $$|2^\mathbb{R}|=|X|=|X^k|=|X^\mathbb{N}|=|X^\mathbb{R}|$$

See a pattern?

It seems that when taking one set to the power of another set, it doesn't really matter how big the exponent is so long as it is smaller than the base. Or thinking about it in terms of the other variable, if the exponent isn't too big then you might as well replace the base by 2. Ie., $$|A| \le |2^B| \Rightarrow |A^B|=|2^B|.$$

Is there a way in which this pattern can be stated and proved rigorously, or is it just a coincidence?

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2 Answers

up vote 6 down vote accepted

If $B$ is infinite, then yes: $|A^B|\leq |(2^B)^B|=|2^{B\times B}|=|2^B|$ while the other direction is trivial.

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Could you elaborate a little bit on that? What is the set $B \times B$ in the formula $2^{B \times B}$ (cartesian product?), and how do we know $|2^{B \times B}|=|2^B|$? If this is too simplistic a reference or link to something explaining it would be great! –  Nick Alger Apr 30 '12 at 8:44
@NickAlger The set $B\times B$ is the Cartesian product, and $|2^{B\times B}|=|2^B|$ since for any infinite cardinal $B$, $|B|=|B\times B|$. You have to be a little careful here showing that bijections between sets correspond to bijections between power sets, etc. though. Some good basic info on Cardinal arithmetic can be found on Wikipedia. –  Alex Becker Apr 30 '12 at 8:53
Thanks, much appreciated. –  Nick Alger Apr 30 '12 at 8:58
The fact that $a^2=a$ for infinite cardinals is shown in answer to this question. –  Martin Sleziak Apr 30 '12 at 9:21
Interesting that it is equivalent to the axiom of choice... I tried to construct a proof just now but was getting stuck so that link was very useful. –  Nick Alger Apr 30 '12 at 9:40
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We must require $A$ to not be $0$ or $1$! In the first case, $|A^B|$ is 0, while in the second case, $|A^B|$ is 1. (Note there are no functions from $B$ to the empty set, and there is one function from $B$ to a set with one element.) Then $|A^B|$ is strictly less than $|2^B|$ when $B$ is $1$ or greater.

Categorize this under "silly but true" :)

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Yes, I realized this and debated whether or not to mention it. I'm always at a loss whether to include little edge case technicalities in my posts. If I do it clutters things up and obscures the main idea. If I don't, someone always comes in and "corrects" me for leaving it out... Thank you for the post though, perhaps it will be useful to others reading in the future. –  Nick Alger May 4 '12 at 10:04
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