# Set of finite and infinite $0$-$1$ sequences countability [duplicate]

I'm trying to prove that a set of a finite sequences of $0,1$ (let's call it $A$) is countable infinite, whereas a set of infinite sequences of $0,1$ (call it $B$) equipotent to $P(\mathbb N)$ is uncountable infinite.

So far I tried showing that the number of sequences possible in A is $\sum \limits_{i=1}^n \ 2^i$. Not sure how to continue from here, if this is even the right direction.

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## marked as duplicate by Srivatsan, robjohn♦, Asaf Karagila, Guess who it is., Arturo MagidinDec 1 '11 at 5:47

I also tried finding some bijective function, but to no avail –  Asaf Nov 30 '11 at 20:06
Use the fact that a countable union of countable sets is countable for the first part. For the other, see here math.stackexchange.com/questions/86650/… –  David Mitra Nov 30 '11 at 20:08
The second part can be understood as a special case of this: math.stackexchange.com/questions/84180/… –  Martin Sleziak Nov 30 '11 at 20:10
I think this is a duplicate of 65988, 61926, and 29599 put together. –  Srivatsan Nov 30 '11 at 20:15
I voted to close it as a duplicate of 29599. Since that question does not cover all parts of this one, I recommend that future voters choose other questions like 61926 or 65988. –  Srivatsan Nov 30 '11 at 21:32

I'll do the first part:

For each positive integer $n$, let $A_n$ be the set of sequences of 0's and 1's of length $n$. Then $A=\bigcup_{n=1}^\infty A_n$.

Moreover, each $A_n$ is finite. In fact $|A_n|=2^n$. Since a countable union of countable sets is countable, it follows that $A$ is countable.

Now the sequence $\underbrace{0\ 0\ 0\ \cdots\ 0 }_{n-\rm terms}\ 1$ is in $A$ for each positive integer $n$. Thus, $A$ is infinite (and so, countably infinite).

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B is a standard application of Cantor's diagonal argument, though the obvious bijection is far more enlightening.

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