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Help with this question would be greatly appreciated. At the moment, I cannot find an f(x) to map these sets.

Also I'm just wondering to solve this problem, since '[]' brackets are used, do the numbers 0 and 1 have to be mapped to from the set (0, ∞), because that really makes the question quite tough, otherwise something like 1/Squareroot(x+1)

p.s. can someone link me the MathJax formatting, cant find it

edit: (0, ∞) is not the reals as it does not go to negative

Thanks!

Jordan Solomons
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If you'd asked for a proof $(0,\,\infty)$ has the same cardinality as $(0,\,1)$, the trick would be easy: specify a bijection as $x\mapsto\frac{1}{\sqrt{1+x}}$ (or whatever choice you prefer; personally, I wouldn't bother with the square-rooting). Your actual question is about $[0,\,1]$, so what do we do? Well, on the one hand we now know $(0,\,\infty)$ can be injected into $[0,\,1]$. On the other hand, the reverse injection is easy (e.g. $x\mapsto x+1$); now finish with the Schröder–Bernstein theorem, which states if two sets can be injected into each other there's a bijection between them. This theorem is used so often in same-cardinality proofs people don't even mention its usage explicitly. Best of all, its proof is constructive, i.e. it tells you how to form the bijection, if you really want one.

J.G.
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  • so essentially (0,1) is the same as [0,1] by a reverse injection x->x+1? I'm having a bit of trouble understanding this, thanks for the help though! – Jordan Solomons Oct 24 '18 at 13:02
  • I think I get it now, so as (0,1) and [0,1] have the same cardinality, and as (0,1) and (0, infinity) also do, therefore [0,1] and [0, infinity] share the same cardinality. Is this the correct way of thinking about this? – Jordan Solomons Oct 24 '18 at 13:16
  • So just to clarify, I was saying $x\mapsto [0,,1]$ injects $[0,,1]$ into $(0,,\infty)$. What I showed is that $(0,,1),,(0,,\infty)$ can be injected into each other, so are the same cardinality; I didn't start by proving $[0,,1]$ is the same cardinality as $(0,,1)$, although of course it is because it injects into a set we've shown is that cardinality. By contrast, if the result of interest was simply that $(0,,1)$ bijects with $[0,,1]$, the latter maps by $x\mapsto (1+x)/3$ into the former. – J.G. Oct 24 '18 at 13:29
  • So by this, can I assume if (0,1) bijects with [0,1], [0,1] also bijects with (0, infinity) or is that a seperate proof. – Jordan Solomons Oct 24 '18 at 13:38
  • @JordanSolomons You can prove it that way round too if you like, since it's obvious that $(0,,1)$ can biject with $(0,,\infty)$. – J.G. Oct 24 '18 at 13:40
  • Mmm, that would be good, too bad I have to use Schroder Bernstein Theorem, I don't think I'll ever figure out this question haha. – Jordan Solomons Oct 24 '18 at 13:48