Prove $[0, 1]$ and $(0, 1]$ are equinumerous

Asked Oct 11 '16 at 20:03

Active Oct 11 '16 at 20:03

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I am having a bit of trobule here. THis post was marked duplicate HOWEVER the open and closed segments were different.

My thought is this:

What my thought was that the bijection was a piecewise function:

$f(x) = 1/(n+1)$ if $x = 1/n$ for some $n \in \Bbb N$ and $x=1/n$ if $x \neq1/n$ for some $n \in \Bbb N$.

However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable $(0,1]$, which is what I tend to think. How could I amend these two functions to get the proper answer?

asked Oct 11 '16 at 20:03

Aggrawal Puja

How does what you have address the issue of $0$? – vadim123 Oct 11 '16 at 20:05
Your definition doesn't make sense. What is $f(.37)$? What is $f(\frac e{\pi})$? Also...could you include the link to the duplicate post you mention? – lulu Oct 11 '16 at 20:05
3

you asked that very same question earlier, it was explained to you why your proposed solution can't work and I showed you a working one, plus the question is a duplicate... – Alessandro Codenotti Oct 11 '16 at 20:06
So couldnt I add f(x) = 1/2 when x = 0? – Aggrawal Puja Oct 11 '16 at 20:07
1

Waiting confidently for the third installment... – Did Oct 11 '16 at 20:27

Prove $[0, 1]$ and $(0, 1]$ are equinumerous

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