Can the unit interval map bijectively to a region?

Question

The Hilbert Curve shows that there exists a surjection from the unit interval to the unit square.

I was wondering, does there exist a bijection from the unit interval to the unit square?

Answer 1

Existence and Construction of bijective map between $(0, 1)$ to $(0, 1)^2$

Yes, there is. The simplest way I can think of constructing one is by invoking the Cantor-Schroder-Bernstein theorem, which tells you that if there are two sets $A$ and $B$, $|A| = |B|$ if $\exists f, g$ such that $$\\f : A \to B \\g: B \to A$$ where both $f$ and $g$ are injective (one-to-one). Note that surjectiveness is not asked for.

We need to construct two maps: $f : (0, 1) \to (0, 1) \times (0, 1)$ and $g: (0, 1) \times (0, 1) \to (0, 1)$ both of which are one-to-one.

$f$ is an easy one. Map a point $a \in (0, 1)$ to the point $(a, a) \in (0, 1) \times (0, 1)$. This is clearly one-to-one. It's not surjective (every point on the square does not have a pre-image) but we don't need one.

Hence, $$f : A \to B, \\f(a) = (a, a)$$

Now, we need $g$. One cute trick is to consider the point $(a, b)$ on the unit square. Let $a$ have the decimal expansion $$a = 0.a_0 a_1 a_2 \ldots$$ and let $b$ have the decimal expansion $$b = 0.b_0 b_1 b_2 \ldots$$

I'll now make a point $c \in (0, 1)$ by taking $$c = 0.a_0 b_0 a_1 b_1 \ldots a_n b_n \ldots$$

Hence, $c$ is the point that you get when you interleave the digits of $a$ and $b$.

So, $$g : B \to A\\ g(0.a_0 a_1 \ldots a_n, 0.b_0 b_1 \ldots) = 0.a_0 b_0 a_1 b_1 a_2 b_2 \ldots $$

$f$ and $g$ give us the two maps we are looking for, hence

$$|A| = |B|$$ or the unit line can be mapped onto the unit square.

Intuition of Cantor Schroder Bernstein

In some sense, having an injective $f : A \to B$ tell you that $|A| \leq |B|$ since you are mapping every point of $A$ uniquely to some point of $B$. Sure, you may have missed points in $B$, but you know that $B$ has, at minimum, the same number of items as $A$.

By symmetry, having an injective $g: B \to A$ implies that $|B| \leq |A|$.

However, since both of these must be true, $$|A| = |B|$$ which is what we were looking for!

The proof is a bit hairy and involves bouncing between infinitely long chains, so I prefer to think of this theorem as a "prove one, use many" times thing.

Answer 2

No. Suppose that yes and let $\phi$ a bijection. Put $Q$ as the square and $I$ the interval.

Now, $Q\setminus\{ \phi(1/2)\}$ is connected but $I\setminus\{ 1/2\}$ not. This is an absurd, because a map convert connected sets into connected sets.

If you want only a function, the answer is yes, the projection $\pi(x,y)=x$ is surjective, and then exist a bijection (because exist a surjection from both sides)