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How do I define a bijection between $(0,1)$ and $(0,1]$?

Establish a one-to-one correspondence between the closed interval $[0,1]$ and the open interval $(0,1)$.

this is a problem in real analysis.


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    $\begingroup$ Hint: taking the domain to be $[0,1]$, define the images of $0$, $1$, $1/2$, $1/3$, $\ldots$ first. $\endgroup$ Aug 3, 2012 at 1:08
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    $\begingroup$ Essentially a duplicate of this. $\endgroup$
    – t.b.
    Aug 3, 2012 at 1:29

2 Answers 2


Possibly you are considering this problem in this textbook but probably not; in any case, the author suggest breaking the proof down into two lemmas that may help you prove this problem:

Part (a): Suppose there are sets $ A,B $ which have a subset $S$ in common and that for some sets $C,D$ we have

  • $A = C \cup S \text{ and } B = D \cup S$
  • $C \cap S = \emptyset \text{ and } D \cap S = \emptyset$
  • there is a 1-1 correspondence between $C,D $

Then use this information to describe a 1-1 correspondence between $A,B$.

Part (b): Describe a 1-1 correspondence between the sets $$\left\{0,1,\frac{1}{2}, \frac{1}{3}, \dots \right\} \text{ and } \left\{ \frac{1}{2}, \frac{1}{3},\frac{1}{4} \dots \right\} $$

Use Parts (a),(b) to prove that the intervals $[0,1]$ and $(0,1)$ are equivalent sets.

  • $\begingroup$ I really appreciate the restraint here, especially from a new user. Good answer, +1 $\endgroup$
    – davidlowryduda
    Aug 3, 2012 at 1:45

I found that using the construction used in the Wikipedia proof of the Schroeder-Bernstein theorem yields an elegant answer.

Take, for instance, $A = [-1,1]$ and $B = (-1,1)$. For the injective map $g:B\rightarrow A$ we can just take the identity, and for $f:A \rightarrow B$ we can use $f(x) = x/2$.

Following through with the construction yields a bijection $\psi: A \rightarrow B$ which is the identity almost everywhere except at points of the form $\pm2^{-n}$.


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