Biject all points on a plane to the real line

Question

I understand that the continuum hypothesis implies (since there are only two infinities; discrete and continuous) that the set of points on an n-dimensional plane is equal to (can be bijected to) the points on a 1 dimensional line. Is there a proof of this for a space of arbitrary dimensionality?

Answer 1

The continuum hypothesis is not what you claim. The continuum hypothesis claims the following:

There is no set with a cardinality that is both strictly greater than $|\mathbb N|$ and strictly smaller than $|\mathbb R|$

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What you are interested is the claim

For all $n\in\mathbb N$, the cardinality of $\mathbb R$ is the same as the cardinality of $\mathbb R^n$.

This claim is true, and is independent from the continuum hypothesis.

The sketch of a proof is such:

First of all, we only need to prove that $|\mathbb R|=|\mathbb R^2|$. This is because, if we prove this, then, from the bijection $f^{[1]}:\mathbb R\to\mathbb R^2$ which maps $x$ to $(f_1(x), f_2(x))$, we can easily construct a bijection $f^{[2]}:\mathbb R^2\to\mathbb R^3$ as follows:

$$f^{[2]}(x, y) = (x, f_1(x), f_2(x)).$$

This is clearly a bijection, and also, it should be clear that using the same idea, inductivelly, we can construct bijections from $\mathbb R^{n}$ to $\mathbb R^{n+1}$ for any $n$.

To prove that $\mathbb R$ and $\mathbb R^2$ have the same cardinality, what we could do is the following mapping: Take some $x\in\mathbb R$. Write it down in decimal form, i.e. $x=a_{2k}a_{2k-1}\dots a_1.b_1b_2b_3\dots$. Now define the mapping that maps this $x$ to the pair $$(a_{2k}a_{2k-2}\dots a_2.b_2b_4b_6\dots, a_{2k-1}a_{2k-3}\dots a_1.b_1b_3b_5\dots)$$

This mapping is almost a bijection. If you just take care of some border cases of infinite nines on the end, you are done.

Answer 2

No, you do not understand the Continuum Hypothesis.

What CH is stating is that there is no cardinality between $\aleph_0$ and $2^{\aleph_0}$. It does not say anything about other cardinalities.

Cantor's theorem tells us that no set is equipotent to its power set, so $$|\Bbb R|<|\mathcal P(\Bbb R)|<|\mathcal P(\mathcal P(\Bbb R))|<\dotsb.$$

And what happens once you've done that through all the natural numbers? Well, then we can take the union and get something even larger. And we can continue.

How far does that go? Well, through the whole backbone of the set theoretic universe. Through the ordinals. There is a reason why the collection of all sets is not a set. It's really really really big.

Now. If $V$ is a vector space over $\Bbb R$ then $|V|=\max\{|\Bbb R|,\dim V\}$. Which means that all the finite dimensional, even $\Bbb R[x]$, and even $\Bbb{R^N}$, all have the same cardinality as $\Bbb R$.

But it also means that once we go beyond $|\Bbb R|$, we get strictly larger vector spaces. So $\Bbb{R^R}$ is strictly larger, since there is an obvious injection from $\mathcal P(\Bbb R)$ into $\Bbb{R^R}$ ($X$ is mapped to its characteristics function). This tells us that its dimension is also very large.

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You could argue that we don't care about these sets all that much. That important sets are sets of natural numbers, or real numbers, or sets of specific vector spaces. And all of those are at most the size of $\Bbb R$. And to some extent, you won't be wrong. It's true that most people only care about such sets. (Even if we omit some obvious counterexamples for sake of argument.)

But then it turns out that the set theoretic universe has an effect about those sets. For example, are projective sets Lebesgue measurable?

So understanding the set theoretic universe is important. And it turns out that it's a whole universe outside the real numbers.