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This is just messing with my head for the past hour.

Question:

$|(0,1)| = |(0,1]|$. If you find a bijection, you don't need to prove that its a bijection.

So a Bijection is one-on-one. E.g $f(a_1) = f(a_2)$ then $a_1=a_2$ and $\operatorname{range} f = B$

In the above case It is bijective is true but is the statement true as well?

and what about if its $|[0,1]^2| = |[0,2]^2|$

Thanks Edit: Heres the exact wording of the question: http://i.stack.imgur.com/2MhxE.png

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    BTW the first part of the question was also poster [here](http://math.stackexchange.com/questions/160738/how-do-i-define-a-$b$ijection-between-0-1-and-0-1).2012-06-20

2 Answers 2

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Edit: In response to your posting of the exact wording of the problem, here is a refined version of my answer.

In the first case, you are trying to show that $(0,1)$ has the same cardinality as $(0,1]$. A bijection $(0,1]\to(0,1)$ that works is $h(x)=\begin{cases}\frac{1}{n+1} & x=\frac{1}{n},\,n\in\Bbb N\\x & \mathrm{otherwise}.\end{cases}$ It is a good exercise to show this is a bijection (even though you don't have to for the assignment).

In the second case, you are trying to show that $[0,1]\times[0,1]$ has the same cardinality as $[0,2]\times[0,2]$. As Arturo points out, the way to start, here, is by finding a bijection $[0,1]\to[0,2]$ (which shouldn't be too difficult). Say that $f$ is such a bijection. Then if we define $g:[0,1]\times[0,1]\to[0,2]\times[0,2]$ by $g(x,y)=\langle f(x),f(y)\rangle$, we have the desired bijection.

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    Excellent point. Fixed.2012-06-20
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Fortunately, an answer to your first question (edited), "Is $|(0,1)| = |(0, 1]|$?" can be found here.

For your second question (also edited), "Is $|[0,1]^2| = |[0, 2]^2|$?", the answer is also "yes": The function $f(x) = 2x$ is obviously a bijection from $[0, 1]\text{ and }[0,2]$ and it's easy to extend this to the product by defining $g:[0,1]^2\rightarrow [0,2]^2$ to be $g((x,y)) = (2x, 2y).$

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    @Cameron Heh. I missed that. $T$hanks.2012-06-21