# How to show $2^{ℵ_0} \leq \mathfrak c$ [duplicate]

I want to show $$2^{ℵ_0}=\mathfrak c$$.

I already showed $$\mathfrak c \leq 2^{ℵ_0}$$ as follows:

Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $$ℵ_0$$ digits. So we have

$$\mathfrak c \leq ℵ_0 * 10^{ℵ_0} \leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$$ since $$ℵ_0 + 4ℵ_0=ℵ_0$$

But how can I prove the other way $$2^{ℵ_0} \leq \mathfrak c$$?

• Look at the Cantor set. Commented Nov 26, 2018 at 13:04

$$2^{\aleph_0}$$ is the cardinality of all reals (belonging to $$(0,1)$$ if you prefer) that you can write by using only $$0,1$$. Those numbers clearly form a subset of $$\mathbb R$$ which must therefore have cardinality at least $$2^{\aleph_0}$$.

You can define a function $$F$$ from the set $$\{ (x_n) | n\in \mathbb{N}, x_n\in \{ 0,1\} \}$$ to $$\mathbb {R}$$ such that $$F[(x_1,x_2,x_3,\ldots )]=0\ .\ x_1x_2x_3\ldots$$
Then this function is injective. So $$\text{Cardinal} \{ (x_n) | n\in \mathbb{N}, x_n\in \{ 0,1\} \} \leq \text{Cardinal}(\mathbb{R})$$ So $$2^{ℵ_0} \leq \mathfrak c$$.

• It’s not 1-1 because $F(0,1,1,\dots)=F(1,0,0,0\dots)$, if I’m not mistaken. But I think the idea is accurate.
– user279515
Commented Nov 26, 2018 at 13:28
• No. It's One to One and your saying is false%% Commented Nov 26, 2018 at 13:32
• Well, I'll let you know that the downvote is mine.
– user279515
Commented Nov 26, 2018 at 13:40
• But why! was my solution false? Commented Nov 26, 2018 at 13:50
• @GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 \dots$ is in binary.
– user279515
Commented Nov 27, 2018 at 15:36