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Question
Consider the intervals of real numbers (0,1) = {x | 0 < x < 1} and (0,1] = {x | 0 < x ≤ 1}.

Show that |(0,1)| = |(0,1]|

Work
I stated that since both intervals are uncountably infinite then they have the same cardinality, but this was marked as wrong. How do I actually prove this?

joe cool
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    There are many, many different sizes of uncountable sets (many, many uncountable cardinals, that is). – BrianO Jun 08 '16 at 05:19
  • So, how do I prove that these two uncountable sets have the same cardinality? – joe cool Jun 08 '16 at 05:23
  • Just because I'm surprised nobody has mentioned this yet - the uncountability of both sets would be enough if you assumed the continuum hypothesis, which is generally not considered acceptable in say, real analysis (it is independent of ZFC and we don't have any real reason to believe in it). – Noah Olander Jun 08 '16 at 05:47

3 Answers3

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First it is easy to show (Using $\tan x$ for instance) that $|\mathbb{R}|=|(0,1)|$. Now it is straight forward to find injections to show that $|(0,1)|\leq |(0,1]|\leq \mathbb{R}$. Now use the Schröder Bernstein theorem to conclude that $|(0,1)| = |(0,1]|$

Ove Ahlman
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1) $(0,1)\subset (0,1] \subset (0,2) $ so $|(0,1)|\le |(0,1]| \le |(0,2)|$. $f:(0,1)\rightarrow (0,2); f (x)=2x$ is a bijection. So $|(0,1)|\le |(0,1]\le |(0,2)|=|(0,1)|$ so $|(0,2)|=|(0,1]|$

2) let $f:(0,1]\rightarrow (0,1)$. $f (1/n)= \frac {1}{n+1}; f (x \ne 1/n)=x $ is a bijection. So $|(0,1)=|(0,1] $.

fleablood
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There are many, many different sizes of uncountable sets — many, many uncountable cardinalities, that is. For example, both $\Bbb R$ and $\{0,1\}^\Bbb R$, the set of all functions from $\Bbb R$ to a 2-element set, are uncountable; but by Cantor's theorem, the latter is larger.

What does it mean for a set to "have a cardinality"? It means that there's a bijection between that set and a cardinal, a particular kind of well-ordered set. (A cardinal is an ordinal which is not bijectible with any smaller ordinal.) Let $\mathfrak{c} = \lvert \Bbb R \rvert$ be the cardinal of the reals.

As Ove Ahlman notes in his answer, the tan function gives a bijection between $(0,1)$ and $\Bbb R$, so there's a bijection $f\colon (0,1) \to \mathfrak{c}$. The finite ordinals $\{0, 1, ..., n, ...\} = \omega$ (essentially, the integers) are an initial segment of $\mathfrak{c}$, there is a bijection $g$ between $\mathfrak{c}$ and $\mathfrak{c} \setminus \{0\}$: $$ g(\alpha) =\begin{cases} f(\alpha) + 1 \quad \text{if $\alpha < \omega$}, \\ f(\alpha) \quad \text{otherwise}. \end{cases} $$ Thus $g\circ f\colon (0,1)\to \mathfrak{c} \setminus \{0\}$ is a bijection. Now extend $g\circ f$ to a bijection $h\colon (0,1]\to \mathfrak{c}$ by defining $h(1) = 0$.

BrianO
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