Proving set cardinality

Question

My question says:

Prove that $(3,6) \cup [10,20)$ has cardinality $\frak{c}$.

Answer 1

We can give an explicit bijection of the set with the reals. The idea is to use that

$$\mathbb{R}=(-\infty,0)\cup [0,\infty).$$ Define

$$(-\infty,0)\to (3,6), x\to \frac{6}{e^x+1}$$ and $$[0,\infty) \to [10,20), x \to \frac{20}{e^x+1}.$$

If you prefer a bijection with $(0,1)$ the idea is the same and the bijection is easier to construct. Write $(0,1)=(0,1/2)\cup [1/2,1).$ Now define

$$(0,1/2)\to (3,6),x\to 6x+3,$$

and $$[1/2,1)\to [10,20), x\to 20 x.$$

Answer 2

Probably the simplest is to say it can't be more than $\mathfrak c$ because it is a subset of $\Bbb R$. Now if you know $(0,1)$ has cardinality $\mathfrak c$, make a correspondence to $(3,4)$, which is a subset of your set, so your set can't be less than $\mathfrak c$. You could make an explicit bijection with $(0,1)$ but that would be a little harder.