How to Prove that $C^{0}$ and $\mathbb{R}$ have equal cardinality ?
$C^{0}$ denote the set of all Continuous function $\mathbb{R} \rightarrow \mathbb{R}$
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Which direction can you do? – Calvin Lin May 27 '13 at 16:15
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1The obvious thing is there are at least as many function as there rare real numbers. (Constant functions in $C^0$) – Mohammad Reza Taesiri May 27 '13 at 16:34
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Hint: A continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ is completely determind by its values on rational numbers $\mathbb{Q}$ (actually in any dense set of $\mathbb{R}$). the cardinality of the range of a function from $\mathbb{Q}$ is at most countable.
tessellation
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3If you take $x\in\mathbb{R}-\mathbb{Q}$ and a sequence $x_n\in \mathbb{Q}$ such that $x_n\rightarrow x$. Then consider $f(x_n)$. This will converge because the function is defined on the whole $\mathbb{R}$. And by continuity the limit will be the value of $f(x).$ – tessellation May 27 '13 at 16:38
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1@GabrielRomon $f(x) = \lim_{h\rightarrow x} f(x) = \lim_{h\rightarrow x , h \in \mathbb{Q} } f(x)$. – Calvin Lin May 27 '13 at 16:38
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Hint: To show that $\mathbb{R} \subset C^0$, consider $f \in C^0$, take $f(0)$.
For the converse,
Hint: $f\in C^0$ is uniquely determined by its values on a dense subset of points.
Hint: $|\mathbb{R^N}| = |\mathbb{R}|$.
Calvin Lin
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