# Prove $\mathbb{R}-\{0\}$ has cardinality $\mathfrak c$

Hint: Prove $(\mathbb{R}-\{0\})\approx\mathbb{R}$, and write $\mathbb{R} =\mathbb{N}\cup (\mathbb{R}-\mathbb{N})$

So far we have shown that $(0,1)$ has cardinality $\mathfrak c$, and all my previous proofs of this sort were made by showing equivalency of a set to $(0,1)$.

I don't understand the hint, and I can't come up with a function from $f: (0,1)\to \mathbb{R} - \{0\}$.

I'd appreciate some pointers.

• $\mathfrak c$ is the cardinality of $\mathbb R$ and also of $(0,1) \subset \mathbb R$. – Mauro ALLEGRANZA Nov 29 '16 at 8:37

## 1 Answer

Consider the map $f(x)$ which is equal to $x$ if $x\in\mathbb{R}-\mathbb{N}^+$, and it is equal to $x-1$ if $x\in\mathbb{N}^+$.

Verify that $f$ is a bijection between $\mathbb{R}-\{0\}$ and $\mathbb{R}$.

P.S. For a bijection between $(0,1)$ and $\mathbb{R}-\{0\}$, take $\cot(\pi f^{-1}(x))$.

• This is like saying... let everything in R except for N stay the same, but shift the naturals back one, that way I pick up the missing zero. So it will be a bijection with R, and since R is an open interval it has cardinality c. – Alex Butterfield Nov 29 '16 at 9:29
• @Alex Butterfield Yes, $f$ is a bijection. Therefore $\mathbb{R}-\{0\}$ has the same cardinatlity of $\mathbb{R}$ that is $\mathfrak c$. – Robert Z Nov 29 '16 at 9:34