$|A|=\mathcal c \ \ |B|=\aleph_0 \ \ A\cap B=\emptyset$ prove that $ |A\cup B|=\mathcal c$

Question

Let $|A|=\mathcal c, \ |B|=\aleph_0, \ A\cap B=\emptyset,$ Prove that $ |A\cup B|=\mathcal c$

So $|A\cup B|=|A|+|B|$ but this just leads to cardinal arithmetic which I don't think is the right approach.

Maybe embedding could work ? There has to be a 1-1 function from $A$ to $(0,1)$ and from $B$ to the naturals, so $A\cup B = \{x\in [0,1], y | x\in \mathbb R, y\in \mathbb N\}$ and now it's enough to know that for all $x$ the cardinality of this set is $\mathcal c$.

Answer 1

Cardinal arithmetic is finding injections. The fact that you write something like:$$\mathfrak c+\aleph_0=\mathfrak c$$ can only be used after you have proved that there is a bijection between two sets, one of cardinality continuum, and another which can be partitioned into sets of sizes continuum and $\aleph_0$.

Of course if you haven't proved these things yet, then you should. But if you have proved some basic things about cardinal arithmetic, then this is a far easier approach: $$\mathfrak c\leq\mathfrak c+\aleph_0\leq\frak c+c=c$$ (Where the latter equality is witnessed by the fact that $[0,1]\cup[2,3]$ is a subset of $\Bbb R$).

Answer 2

Your embedding shows (as $A\cup B \subseteq \mathbb R$) in this case, $|A\cup B| \le \mathfrak c$. As $\mathfrak c \le |A\cup B|$ (since $A \subseteq A \cup B$), the statement follows by Schröder-Bernstein.