Please help me prove the following equalities between set cardinalities by explicitly showing an appropriate mapping:

$$\left | (0,1) \right |= \left | (1,+\infty ) \right |$$

  • $\begingroup$ If you have multiple related questions, please put them together in one post! Also, show us the steps you've already taken and where you're stuck $\endgroup$ – konewka Oct 16 '14 at 9:31
  • $\begingroup$ And you might want to consider the $\tan$-function or something containing $\frac{1}{x}$ for this one $\endgroup$ – konewka Oct 16 '14 at 9:32
  • $\begingroup$ Thank you for your comments, I will have a go with tan or 1/x $\endgroup$ – Will Oct 16 '14 at 9:34
  • $\begingroup$ I'm not sure if I understand your last comment correctly, but the functions $\tan(x)$ and $\frac{1}{x}$ are not directly bijections between those two intervals, you'll have to work with them a little more $\endgroup$ – konewka Oct 16 '14 at 9:42

The function $f(x)=1/x$ will suffice to form a bijection between $(0,1)$ and $(1, \infty)$. I'm not sure how much you know about comparing cardinalities, but the standard method is to form a bijective function between the two sets to be compared. Is it clear to you how this function does that?

  • $\begingroup$ Thanks for your answer, yes this helped out. I have only just started seeing cardinalities and am still slightly confused but I believe this makes sense. $\endgroup$ – Will Oct 16 '14 at 9:43
  • $\begingroup$ That's very normal, and you're not alone! Glad I could help. If you want a little extra practice/challenge, try showing that the set of all roots of polynomial equations with rational coefficients (the algebraic numbers) is countable (i.e. the same cardinality as $\mathbb{N}$). $\endgroup$ – epsilonics Oct 16 '14 at 9:50

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