# Find a one-to-one correspondence between $[0,1]$ and $(0,1)$. [duplicate]

Possible Duplicate:
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.

Thanks

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## marked as duplicate by robjohn♦, Pedro Tamaroff♦, MJD, William, Rudy the ReindeerAug 3 '12 at 6:33

Hint: taking the domain to be $[0,1]$, define the images of $0$, $1$, $1/2$, $1/3$, $\ldots$ first. – David Mitra Aug 3 '12 at 1:08
Essentially a duplicate of this. – t.b. Aug 3 '12 at 1:29

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.

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I really appreciate the restraint here, especially from a new user. Good answer, +1 – mixedmath Aug 3 '12 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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