# Cardinality of a union of uncountable and countable set.

Suppose you have a set A which has the same cardinality with real numbers R, which means |A| = |R|. Also suppose that you have a finite set B , which of course has finite cardinality. Also suppose A is a proper subset of A union B, I mean there are some objects which are not an element of A but they are an element of B.

Now is the cardinality of A union B greater than the cardinality of real numbers, also if so is this cardinality of A union B equal to cardinality of the power set of real numbers ?

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math.stackexchange.com/questions/89104/… I think it has cardinaliy of real numbers is it true ? –  mehdi Jul 24 '12 at 21:52
This question is very elementary, but I wouldn't call it bad -- why the downvotes, with no explanatory comments? –  user7530 Jul 24 '12 at 22:42

The question, in a nutshell, asks whether or not $n+\frak c=\frak c$, for finite $n$, where $\frak c$ is the cardinality of the continuum.

The answer is yes, if we add to the continuum a finite set then we do not increase its cardinality. In fact even if we add a set of size continuum, namely $\frak c+c=c$, and even $\frak c\times c=c$.

To reach the cardinality of ${\cal P}(\Bbb R)$ we need a fairly large set, i.e. $A+{\frak c}={2}^\frak c$ then $|A|=2^\frak c$, and this can be pretty large in ZFC. In fact, it can be just about any cardinal which has a cofinality of $<\frak c$.

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The cardinality of the union is exactly the cardinality of $\mathbb R$. To see that, take the extreme case that $A$ and $B$ are disjoint and use the bijections $f:A\to R$ and $g:B\to \{0,\dots,k-1\}$ where $k=\lvert B\rvert \in \mathbb{N}$ to construct a bijection $\phi: A\cup B\to\mathbb{R}$ as follows: $$\phi(x) = \cases{ g(x) & for x\in B\\ f(x)+k & for x\in A and f(x)\in\mathbb{N}\\ f(x) & for x\in A and f(x)\notin\mathbb{N} }$$ If $A\cap B\neq\emptyset$ then use the fact that $A\cup B=A\cup(B\setminus A)$ and do the same construction with $B\setminus A$ instead of $B$ (of course, then $k=\lvert B\setminus A\rvert$).

Note that a similar construction works also if $B$ is countable infinite. In that case, map $B$ to the odd numbers and replace $f(x)+k$ with $2f(x)$ for $f(x)\in \mathbb N$.

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+1. Note that the result holds even if $B$ has the same cardinality as $\mathbb R$. –  Alex Becker Jul 24 '12 at 22:19
@AlexBecker: Yes, however then one needs a different strategy to make place for uncountably many elements; functions mapping the reals to the negative and to the non-negative numbers are an obvious choice. –  celtschk Jul 24 '12 at 22:25