Prove that the set of all real functions defined on the closed unit interval [0,1] has cardinal number $2^c$. it is easy to see that there exists as many as such functions i.e. the characteristic functions but I cannot prove the exact cardinality. so comments will be helpful
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2Any function can be identified with its graph, so the cardinality is at most that of the power set of $[0,1]\times\mathbb R$, i.e. $2^c$ – Etienne Aug 28 '13 at 16:33
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1Use the fact that $\mathfrak c^\mathfrak c=2^\mathfrak c$ and $[0,1]\sim \Bbb R$. – Git Gud Aug 28 '13 at 16:34
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Just below my comment above it says 'Trivial answer converted to comment'. Can anyone else see that? – Git Gud Aug 28 '13 at 16:35
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@AsafKaragila Thanks. It disappeared to me too as soon as I posted my question. – Git Gud Aug 28 '13 at 16:36
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it is true that $c^c=2^c$, but I want to prove this @Git Gud – SHIBASHIS Aug 28 '13 at 16:39
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1@SHIBASHIS Then just open the link above. – Git Gud Aug 28 '13 at 16:40
2 Answers
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HINT: Use the fact that $|A^B|=|A|^{|B|}$, and show that $|\Bbb{R^R}|=2^{|\Bbb R|}$.
Asaf Karagila
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To expand on Etienne's comment, we have the following argument:
We have an injection from $\mathcal P([0,1])$ to the set of real functions on $[0,1]$ (which I will call $S$ for convenience) since each subset of $[0,1]$ can be mapped to a unique corresponding characteristic function.
On the other hand, we have an injection from $S$ to $\mathcal P([0,1]\times \mathbb R)$ since each real function in $S$ can be mapped to its unique graph in $[0,1]\times\mathbb R$.
From there, we have only to show that $\left|[0,1]\right|=\left|[0,1]\times\mathbb R\right|$
With the above, we have $$ \left|\mathcal P([0,1])\right|\leq \left| S\right| \leq \left|\mathcal P([0,1]\times \mathbb R) \right|=\left|\mathcal P([0,1]) \right| $$ Thus, $\mathcal P([0,1])$ and $S$ have equal cardinalities.
Ben Grossmann
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