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If you know that a given set O is countable.

$\#O\leq \#\mathbb{N}$

Does this imply that the following statement holds?

$\# O \leq \#\mathbb{R}$

I'm not sure, but I think it makes sense, because $\mathbb{N}\subset\mathbb{R}$ and therefore you can construct easily an injective function between $\mathbb{R}$ and $\mathbb{N}$.

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up vote 3 down vote accepted

Yes, it does, and your reasoning is correct: you have an injection $f:O\to\Bbb N$, and the map $g:\Bbb N\to\Bbb R:n\mapsto n$ is an injection, so the composition $g\circ f:O\to\Bbb R$ is also an injection.

In fact it implies that $\# O < \#\mathbb{R}$, since $\Bbb R$ is uncountable.

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Isn't it more to the point to say $\mathbb{R}$ has countable subsets than that $\mathbb{R}$ is uncountable? If a set has countable subsets, the the cardinality of every countable set is less than the cardinality of that set. – Michael Hardy Sep 6 '12 at 19:08
@Michael: Are you using "less than" to refer to "$\le$"? With "$<$" it's not true. – Trevor Wilson Sep 6 '12 at 19:13
@Michael: My point was that although the argument given merely shows that $|O|\le|\Bbb R|$, a stronger statement is in fact true. In other words, my second paragraph was a separate observation. (And what you say isn’t right: all you get is less than or equal to, not less than.) – Brian M. Scott Sep 6 '12 at 19:15
Sorry---I should have said $\le$. – Michael Hardy Sep 7 '12 at 3:04

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