How to prove that $2^\omega=\mathfrak{c}$?

Let $\mathfrak{c}$ denote the continuum.

My textbook says that $2^\omega=\mathfrak{c}$. How can one prove this equality?

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Prove that there is an injection from $2^{\omega}$ to $(0,1)$; then prove that there is an injection from $(0,1)$ to $2^{\omega}$; then invoke Cantor-Bernstein-Schroeder. Finally, find a bijection between $\mathfrak{c}$ and $(0,1)$. –  Arturo Magidin Jul 20 '12 at 3:20
Thanks. However the difficult thing for me, is looking for the injections from eachother. –  Paul Jul 20 '12 at 3:26
To inject $(0,1)$ to $2^{\omega}$, consider binary expansions. To inject $2^{\omega}$ to $(0,1)$, identify $2^{\omega}$ with $P(\mathbb{N})$, and think about indicator functions. –  Arturo Magidin Jul 20 '12 at 3:28
The tag [ordinals] implied $2^\omega$ is ordinal exponentiation, which is $\omega$... –  Asaf Karagila Jul 20 '12 at 4:39
Sorry, Arturo Magidin, I still cann't catch you. –  Paul Jul 20 '12 at 5:33
Clearly $2^\omega$ has cardinality equal to $|P(N)|$. Think of a subset of the naturals as a sequence of zero's and one's. (A one indicates a particular element is in the subset, a zero indicates it isn't). Between every digit, put a 3 and in the begining put a decimal point. (Eg $11001$... becomes $0.131303031$...; this is done to overcome the problem of ambiguity in representations)
Now every subset has been put in a one one correspondence with a real number. So $|P(N)|\le |(0,1)|$. Conversely every real number between $0$ and $1$, has a binary expansion albeit sometimes redundantly; and removing the point gives us a 0-1 sequence, i.e. a subset of the naturals; and hence that particular set of reals has cardinality $\le |P(N)|$. By Schroder Bernstein theorem we have $|P(N)|=|(0,1)|$. It is now trivial to conclude that $|P(N)|=|R|$ (Hint: Think of the tan function.) As an added bonus by Cantor's theorem ($|A|<|P(A)|$) we have also established that the reals are uncountable.