real analysis - Is there a bijection between $mathbb{R}^2$ and $(0,1)$?

True or false: there exists a bijection between $\mathbb{R}^2$ and the open interval $(0, 1)$.

I think this is false, because $R^2 - \{0\}$ is connected but $(0,1) = R$ as $R-\{0\}$ is not connected, as the continuous image of a connected set is connected.

Answer

The answer is True.

At this time I cannot provide the explicit bijection BUT the idea is this:

Schroder–Bernstein Theorem:

Assume there exists a $1–1$ function $f : X \rightarrow Y$ and another $1–1$ function $g : Y \rightarrow X$. Then there exists a $1–1$, onto function $h : X \rightarrow Y$ and hence $X ∼ Y$ .

Define $f:(0,1) \rightarrow (0,1)\times (0,1)$ by $$f(x)=(x,1/3)$$
Then $f$ is injective.

Define $g:(0,1)\times (0,1) \rightarrow (0,1)$ by $$g(x,y)=0.x_1y_1x_2y_2....$$
where $x=0.x_1x_2x_3....$ and $y=0.y_1y_2y_3....$ and where we make the convention that we always use the terminating form over the repeating $9's$ form when the situation arises.

Then $g$ is injective. (Prove this!)

Hence $$(0,1) \sim (0,1) \times (0,1)$$

We know $(0,1)$ and $\Bbb{R}$ are equivalent,via $$x \mapsto \tan \pi (2x-1)/2$$

So, next map $\Bbb{R}^2$ bijectively onto the open unit square $(0, 1)\times (0,1)$ by mapping each $\Bbb{R}$ bijectively onto the open interval $(0, 1)$

Hence $$(0, 1)\times (0,1) \sim \Bbb{R}^2$$

Summary:

$$(0,1) \sim (0,1) \times (0,1) \sim \Bbb{R}^2$$

In addition, if your function is continuous, then this is not true!