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I have a hard time doing this proof. Can anyone help me?

Show that the following pairs of sets S and T are equinumerous by finding a specific bijection between the sets in each pair.

$S = [0,1]$ and $T = [0,1)$

Itsnhantransitive
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  • Can you find a bijection between ${0,1,2\ldots}$ and ${1,2,\ldots}$? – John Griffin Sep 29 '17 at 01:23
  • ... and then, can you extend the idea to find a bijection between ${1, 1/2, 1/4, \dots}$ and ${1/2, 1/4, \dots}$? Then the bijection between $S$ and $T$ can be the identity map on the complements of these two sets. –  Sep 29 '17 at 01:31
  • I think I got the proof done! Thank you! – Itsnhantransitive Sep 29 '17 at 01:33

1 Answers1

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Well, you've probably seen that the identity map won't work because isn't a bijection. The $1$ in the set $S$ can't be mapped to $1$ in the set $T$, because $1 \notin T$.

So we have the identity map but with the caveat that we try to map the $1$ in $S$ to something else. Say we send it to $\frac{1}{2}$. This causes a new problem, obviously. Lets send the $\frac{1}{2}$ to $\frac{1}{3}$. This causes another new problem. Continuing in this manner, we end up with the identity map for most of the elements of $S$, but we're doing something different to a small (but countably infinite) subset of $S$.

Namely, we are mapping $\frac{1}{n}$ to $\frac{1}{n+1}$ for $n > 0$ (where $n \in \mathbb{Z}$).

I'll leave it to you to get this down into proper maths, but I think I've given you the gist of the argument that you need to make.

Matt
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