# Is a countable set $A$ dense in an uncountable set $B$ by analogy of $\mathbb{Q}$ being dense in $\mathbb{R}$?

Let $A$ be any countable set and $B$ any uncountable set with the same infinity as $\mathbb{R}$. Then we must have a one-to-one map $\phi:A\rightarrow\mathbb{Q}$. So, can we say by all this that $A$ is in fact dense in $B$, since it is well known that $\mathbb{Q}$ is dense in $\mathbb{R}$ ?

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When we say $\mathbb{Q}$ is dense in $\mathbb{R}$, we're always refering to a back ground (partial) order (or topology). For a generic uncountable set $B$, we don't have a notion of order or topology, hence no notion of denseness. You must first choose a partial order or topology on $B$, then you can ask if a given $A$ is dense. –  Jason DeVito Aug 20 '12 at 12:43
Does "the same infinity" mean "the same cardinality"? –  rschwieb Aug 20 '12 at 12:44
@rschwieb yes I think so - I'm referring to the different types of infinities possible, i.e. whilst the natural numbers contain an infinite number of elements, they are countable, whereas the reals are not countable. –  picasso Aug 20 '12 at 12:51

No.

$\mathbb{Z}$ and $\mathbb{Q}$ have the same cardinality, but one is dense in $\mathbb{R}$ and the other is not.

Added I also wanted to say that "denseness" of a subset is not a measure of how many points there are in the subset, it is a measure of how uniformly they are spread throughout the superset and how close they get to points of the superset. In the usual topology on $\mathbb{R}$, every point of $\mathbb{R}$ is approached by points of $\mathbb{Q}$, but you cannot approach all points of $\mathbb{R}$ with points from $\mathbb{Z}$. The gaps in $\mathbb{Z}$ are just too big.

In the other answers you can read that "topology" determines when the subset is spread out enough that it "gets close" to every point of the superset.

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Density requires a notion of topology.

Without specifying the topology on the set $B$ you cannot decide whether or not $A$ is dense.

For example, if we take a set of the same size as $\mathbb R$, and consider the co-finite topology, namely open sets are those whose complement is finite. In such topology every countably infinite set is dense.

Generally speaking, though, it does not hold for every topology, as the other answers demonstrate.

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No this is not even true even if you're working in $\Bbb{R}$. Take $A = \{1,1/2,\ldots, 1/n,\ldots\}$ and $B = \Bbb{R}$.

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No. $\bf N$ is a countable set and $\bf R$ has the same cardinality as $\bf R$ (surprisingly ;) ), but $\bf N$ is certainly not dense in $\bf R$.

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Note that one can go to the almost opposite extreme as in Asaf's example above, and consider $\mathbb{R}$ with the discrete topology, so that all subsets are open.

Then the only dense set in this topological space is the entire set $\mathbb{R}$ itself. (So, in particular, no countable subset is dense.)

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In fact co-countable topology is enough to ensure that all countable sets are closed and nowhere dense... :-) –  Asaf Karagila Aug 20 '12 at 13:07