More numbers between $[0,1]$ or $[1,\infty)$?

Solution 1:

The sets $[0,1]$ and $[1,\infty)$ have the same cardinality (i.e., there are exactly as many numbers in one as in the other). You can construct an explicit bijection between them with some tinkering on the tangent function (which provides a bijection between $[0,\pi/2)$ and $[0,+\infty)$).

In fact, you can easily construct a bijection between $[0,1)$ and $[1,\infty)$, which I suppose is sufficient for you at this point. It is more complicated to do it from $[0,1]$, see for example this question.

Solution 2:

This isn't an answer to your exact question, but I think this picture really helped my intuition about this question:

On top we have a circle centered at, say, $(0,2)$ with radius one. Extend a radius through the circle until it hits the $x$-axis and we'll always get two points: $x$ and $x'$, in the graph.

Tracing out from left to right, we trace out both the semicircle and the whole $x$-axis, i.e. the real line.

But the former only has length $\pi$; with some rigor we can show that this means there are exactly as many numbers in and subset of the reals as there are reals.

Solution 3:

The accepted answer is correct and answers your question perfectly.

You might read later in your education the sense in which $A:=[0,1]$ is almost finite while $B:=[1,\infty)$ is not. This almost finiteness is known as compactness and sets that are compact are in a different way smaller than non-compact sets.

For example, given any $\varepsilon>0$, you can quite easily find a finite number of elements of $A$, $\mathcal{C}:=\{x_1,x_2,\dots,x_N\}$ such that for all $x\in A$ there exists an $x_i\in \mathcal{C}$ such that

$$|x-x_i|<\varepsilon.$$

This is not the case for $B$.