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

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

Thanks ahead:)

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    Sorry, Arturo Magidin, I still cann't catch you.2012-07-20

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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.