I know that the set of irrational numbers $\mathbb{P}$ is dense. So I can establish the following set: \begin{equation} A=\{\mathbb{P}~\cup~\{x\}~|~x~\in~\mathbb{Q}\} \end{equation} As i know that $\mathbb{Q}$ has cardinality $\aleph_0$, then this set A should have this same cardinality but this is not the case. Where am I wrong?
How to prove that $\mathbb{R}$ has $2^{c}$ uncountable dense subsets.
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general-topology
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1It's $2^{\frak c}$, not $2^{\aleph_1}$. The size of $\mathbb R$ is not $\aleph_1$ unless you assume continuum hypothesis. – 2017-01-28
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0Set $A$ is countable. It does not contain irrationals and there are no subsets of cardinality $2^{\aleph_1}$ – 2017-01-28
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6For every subset $B\subseteq \Bbb P$, the set $B\cup\Bbb Q$ is dense. – 2017-01-28
2 Answers
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$\mathbb{P}$ has size $\mathfrak{c}$, so it has $2^\mathfrak{c}$ many subsets.
For every subset $S \subseteq \mathbb{P}$ define $D(S) = S \cup \mathbb{Q}$. This is dense as it contains $\mathbb{Q}$ and there are $2^\mathfrak{c}$ sets of the form $D(S)$, all different.
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0I was almost finished writing this same argument up when I saw yours! +1 – 2017-01-28
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The title suggests the problem is to show there are $2^c$ dense subsets of $\mathbb R,$ each of which is uncountable. To obtain this we should remark that the cardinality of the set $X$ of finite or denumerable subsets of $\mathbb P$ has the cardinality of $\mathbb R.$ Thus $\mathcal P( {\mathbb P}) \setminus X$ has the cardinality of $\mathcal P( {\mathbb P}),$ namely $2^c.$ So the sets $E \cup \mathbb Q, E \in \mathcal P( {\mathbb P}) \setminus X,$ do the job.