# Proof that $\mathbb{R}^2 \cong \mathbb{R}$

I'm trying to prove that $$\mathbb{R}^2$$ and $$\mathbb{R}$$ have the same cardinality. Here is my attempt. I will be taking the Schröder-Bernstein theorem for granted.

Define the map $$f: \mathbb{R} \to \mathbb{R}^2$$ sending $$x \longmapsto (x,0)$$. I claim that $$f$$ is an injection. Indeed, if $$f(x) = f(y)$$ for $$x,y \in \mathbb{R}$$, then $$(x,0) = (y,0)$$, so $$x = y$$. Therefore, $$|\mathbb{R}| \leq |\mathbb{R}^2|$$. By the Schröder-Bernstein theorem, it suffices to prove that $$|\mathbb{R}^2| = |\mathbb{R}|$$.

Lemmma 1: $$\mathbb{R} \cong (0,1)$$.

Proof of Lemma 1. Notice that $$\tan(x)$$ is a bijection from $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}$$ since $$\tan$$ has period $$\pi$$; furthermore, the existence of the $$\arctan$$ function guarantees that the function is surjective, hence bijective. Notice that there is a one-to-one correspondence $$\left(0,1\right) \to \left(- \frac{\pi}{2}, \frac{\pi}{2}\right)$$ sending $$x \longmapsto \left(x - \frac{1}{2}\right)\pi$$. Composing bijections gives a bijection $$\left(0,1\right) \to \mathbb{R}$$.

Lemma 2: Given sets $$A, B, C,D$$ such that $$A \cong C$$ and $$B \cong D$$, we have $$A \times B \cong C \times D$$.

Proof of Lemma 2. Since $$A \cong C$$ and $$B \cong D$$, there exist bijections $$f: A \stackrel{\sim}{\to} C$$ and $$g: B \stackrel{\sim}{\to} D$$. Then define the map $$h: \underset{(a,b)}{A \times B} \underset{\longmapsto}{\to} \underset{(f(a), g(b))}{C \times D}.$$ Given $$(a_1, b_1), (a_2, b_2) \in A \times B$$ for which $$h(a_1, b_1) = h(a_2, b_2)$$, we have $$(f(a_1), g(b_1)) = (f(a_2), g(b_2))$$. So $$f(a_1) = f(a_2)$$ and $$g(b_1) = g(b_2)$$. The former implies $$a_1 = a_2$$ by injectivity of $$f$$, while the latter implies $$b_1 = b_2$$ by injectivity of $$g$$. So $$(a_1, b_1) = (a_2, b_2)$$, so $$h$$ is injective. Given $$(c,d) \times C \times D$$, by surjectivity of $$f$$, there exists $$a \in A$$ such that $$f(a) = c$$; by surjectivity of $$g$$, there exists $$b \in B$$ such that $$g(b) = d$$. We then have $$(c,d) = (f(a), g(b)) = h(a,b),$$ so $$h$$ is surjective, hence bijective, so $$A \times B \cong C \times D$$.

Corollary. It suffices to prove that $$(0,1) \times (0,1) \cong (0,1)$$.

Proof. We know $$\mathbb{R} \cong (0,1)$$ by Lemma 1. By Lemma 2, that implies that $$\mathbb{R} \times \mathbb{R} \cong (0,1) \times (0,1)$$. So if we prove $$(0,1) \times (0,1) \cong (0,1)$$, we have $$\mathbb{R} \times \mathbb{R} \cong (0,1) \times (0,1) \cong (0,1) \cong \mathbb{R},$$ which is the intended result.

Claim. $$(0,1) \times (0,1) \cong (0,1)$$.

Proof of Claim. We use the Schröder-Bernstein theorem. There is an obvious injection $$(0,1) \times (0,1) \times (0,1)$$ sending $$x \longmapsto \left(x, \frac{1}{2}\right)$$. This is an injection regardless of whether we allow an infinite string of $$9$$'s. Concordantly, we can define an injection $$(0,1) \times (0,1) \to (0,1)$$ as follows. Given $$(x,y) \in (0,1) \times (0,1)$$, choose a decimal expansion for $$x,y \in (0,1)$$ that does not contain an infinite string of nines, which we know to be unique. Write $$x = 0.x_1 x_2 x_3 \ldots \\ y = 0.y_1 y_2 y_3 \ldots$$ Then define $$f(x,y) = 0.x_1 y_1 x_2 y_2 x_3 y_3 \ldots$$ Since neither $$x,y$$ contained an infinite string of nines, $$f(x,y)$$ likewise does not. I claim that $$f$$ is an injection. Indeed, suppose that $$f(a,b) = f(c,d)$$ for $$(a,b), (c,d) \in (0,1) \times (0,1)$$. Defining the decimal expansions for $$a,b,c,d$$ similarly, we then have $$0.a_1 b_1 a_2 b_2 a_3 b_3 \ldots = 0.c_1 d_1 c_2 d_2 c_3 d_4 \ldots.$$ Since neither of these expressions contain an infinite string of $$9$$'s, we can equate corresponding digits, so $$a_i = c_i$$ for all $$i \geq 1$$ and $$b_i = d_i$$ for all $$i \geq i$$, so $$a = c$$ and $$b = d$$, so $$(a,b) = (c,d)$$, so $$f$$ is injective. By the Shroder-Bernstein theorem, $$(0,1) \times (0,1) \cong (0,1)$$.

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How does this proof look? The only fact I am not totally sure I understand fully is why the decimal expansion not terminating in a string of nine's is unique. I believe we can assert that every element of $$(0,1)$$ has exactly two decimal expansions. What is the canonical way to go about showing this?

• $x\mapsto (x,1/2)$ is not an injection $(0,1)^2\to (0,1),$ but the other way, $(0,1)\to(0,1)^2.$ May 8, 2021 at 4:39
• Sorry, this was a typo. I will fix it now. May 8, 2021 at 4:40
• Also, $(0,1)\to(0,1)\to(0,1)$ maybe should be $(0,1)\times (0,1)\to (0,1)?$ Otherwise, the proof looks fine. Very well-organized. May 8, 2021 at 4:42
• This looks fine. At some point soon you'll learn that for any infinite set $A, \vert A \times A \vert = \vert A \vert$. That will simplify matters a lot. May 8, 2021 at 4:42
• A real number has at most two decimal expansions as you said, and furthermore the only ones having two different expansions are real numbers of the form $n/10^{-m}$ where $n, m$ are integers. You can find this result as an exercise on Appendix B of Terence Tao's Analysis I. It is also proven in detail in the book "The number systems" of Solomon Feferman which I think it is really good. May 8, 2021 at 11:07

The map $$(0,1)\times(0,1)\to(0,1)$$ you build is certainly injective, but it is not bijective. Take $$0.191919\dotsb$$ to see why. However you don’t need surjectivity.

• He doesn't need it to be a bijection, as he's using Schröder-Bernstein theorem. May 8, 2021 at 10:29
• @Eparoh True but he says that it is which is a mistake. May 8, 2021 at 10:48
• True, I didn't notice it. I think it should be a typo because after all the work, it seems he knows what he's applying. May 8, 2021 at 11:02
• @Eparoh Maybe or maybe not since he admits confusion on decimal representations. Let's wait to hear from him. May 8, 2021 at 11:07
• This was indeed a typo. Thank you for pointing that out. May 8, 2021 at 17:32

Egreg has identified a problem but you might like more detail.

Your map $$f$$ from $$(0, 1) \times (0, 1) \rightarrow (0, 1)$$ is an injection but not a bijection, see below. However, if it was a bijection then you would not need your map the other way or Schröder-Bernstein since the bijection, by itself, would show that the sets have the same cardinality.

You say that every element of $$(0, 1)$$ has exactly two decimal expansions. Most have just one. It is only the ones that end in all $$9$$s or all $$0$$s that do. See this previous question for details.

Now, I say that your map is an injection but not a bijection. So, I am saying that it is not surjective. I need an example to prove this. What maps to $$0.459090909...$$? It's $$(0.4999..., 0.5000...)$$ except the representation $$0.4999...$$ breaks your rule, it is the wrong representation of $$0.5000...$$. So that point is in fact $$(0.5000..., 0.5000...)$$ which maps to $$0.55000...$$.

However, this does not destroy your proof. It just means you need Schröder-Bernstein after all. You have an injection both ways. The whole point of Schröder-Bernstein is that it avoids the need to establish an explicit bijection which is often messy or difficult.