How to prove that the Torus and a Ball have the same Cardinality

Question

How to prove that the Torus and a Ball have the same Cardinality ? The proof is using the Cantor Berenstein Theorem. I now that they are subsets of $\mathbb{R}^{3}$ so I can write $\leq \aleph$ but I do not know how to prove $\geqslant \aleph$. Thanks

Answer 1

Hint:

Show that a circle is equipotent with $[0,2\pi)$ by fixing a base point, and sending each point on the circle to its angle, where $0$ is the base point.

Since both the torus and the ball contain a circle, both have at least $\aleph$ many elements.

Answer 2

I'm assuming that you're talking about solid balls and tori in $\mathbb R^3$.

To prove that these sets are equipollent using the Cantor–Bernstein–Schröder theorem, all you need to do is to show that:

1. there exists an injection $f$ from the ball $B$ into the torus $T$, and
2. there exists an injection $g$ from the torus $T$ into the ball $B$.

The CBS theorem then says that there exists a bijection between $B$ and $T$.

If you're given explicit geometric definitions of $B$ and $T$, you can use these directly to construct invertible affine maps $f$ and $g$ such that $f(B) \subset T$ and $g(T) \subset B$. Another, more general way to do this is to note that both of these subsets of $\mathbb R^3$ are bounded and have non-empty interior. Thus, for each of them, there exists open balls $I$ and $O$ such that $I_B \subset B \subset O_B$ and $I_T \subset T \subset O_T$. Then choosing $f$ and $g$ (again as affine maps) such that $f(O_B) = I_T$ and $g(O_T) = I_B$ will do the trick.

Answer 3

Let's assume known the fact that $\Bbb R^m$ and $\Bbb R^n$ have the same cardinality for all pairs of positive integers $(m,n)$.

Then the following follows easily: Let $X\subset\Bbb R^n$ such that there exists a subset $A\subset X$ homeomorphic to an open subset of $\Bbb R^m$ for some $m\leq n$. Then $|X|=|\Bbb R^n|$.

Indeed, such an $A$ obviously contains an open segment $I$ (e.g. any radius of any open ball contained in $A$) and the chain of inclusions $$ I\subset A\subset X\subset\Bbb R^n $$ together with the fact that $|I|=|\Bbb R|=|\Bbb R^n|$ shows that $|X|=|\Bbb R^n|$.

It is clear that both the ball $B$ and the torus $T$ (no matter whether you take the solid shapes or just their surfaces) satisfy the property required for $X$, thus $|B|=|\Bbb R^3|=|T|$.