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I need to show that the two sets $[0,1]$ and $[0,1)$ have the same cardinality. I know that in order to show this I must show that there exists $f$ such that $f:[0,1]\to[0,1),$ but I am not sure how to proceed.

Any help would be appreciated. Thanks.

Martin Sleziak
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Mark Watson
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3 Answers3

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Define:

\begin{equation} f(x)=\begin{cases} \frac{1}{1+n}, & \text{if $x = \frac{1}{n}$ , $n \in \mathbb{N}$ }.\\ x , & \text{otherwise}. \end{cases} \end{equation}

gamma
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These sorts of things often boil down to using the fact that there is a bijection between the nonnegative integers and the positive integers, allowing you to shift everything over by one to make room for an extra point (e.g. the Hilbert hotel paradox); you just need to find a suitable copy of $\mathbb{N}$ in the problem.

In fact, we can completely characterize the bijections:

Lemma: every bijection between $[0,1]$ and $[0,1)$ can be uniquely expressed as:

  • An injection $g : \mathbb{N} \to [0,1)$
  • A bijection $h : [0,1) \setminus g(\mathbb{N}) \to [0,1) \setminus g(\mathbb{N})$

and conversely, given such data, there there is a corresponding bijection

$$ f : [0,1] \to [0,1) : x \mapsto \begin{cases} g(0) & x = 1 \\ g(g^{-1}(x) + 1) & x \in g(\mathbb{N}) \\ h(x) & x \in [0,1) \setminus g(\mathbb{N})\end{cases} $$

  • This is really the same spirit as my answer... – Ian Sep 27 '15 at 23:29
  • @Ian: Well, every answer is ultimately going to boil down to "write $[0,1) \cong A \cup \mathbb{N}" so there is a limit on how much variation is available. I find my presentation somewhat more transparent, otherwise I wouldn't have bothered posting. –  Sep 27 '15 at 23:38
  • @Ian Indeed. But I do like the intuition presented in the first paragraph. Especially for the level of the question, when these concepts can seem bizzare. A variety of presentations of the same notions is a good idea: different people will find different presentations more in tune with their way of thinking. – Selene Routley Sep 28 '15 at 01:39
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    @WetSavannaAnimalakaRodVance I like it better now that the first paragraph has been edited in. – Ian Sep 28 '15 at 01:41
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Suppose $A \subset [0,1]$ is countably infinite and $1 \in A$. Then $[0,1] \setminus A=[0,1) \setminus A$. So the identity function, call it $f$, is a bijection from $[0,1] \setminus A$ to $[0,1) \setminus A$. Can you define a bijection, call it $g$, from $A$ to $[0,1) \cap A$? Once you have done that,

$$h(x)=\begin{cases} g(x) & x \in A \\ f(x) & x \not \in A \end{cases}$$

is a bijection from $[0,1]$ to $[0,1)$. A hint for constructing $g$: consider enumerating $A$ as a sequence $\{ x_n \}_{n=1}^\infty$ with $x_1=1$.

Ian
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