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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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A bijection is more than one-to-one: it must also be onto.... You say "the statement is true, but is the statement true as well?" I don't understand what you are asking. If by $(0,1)^2$ and $(0,2)^2$ you mean the sets of all pairs of reals in the original interval, then, yes, $|(0,1)^2| = |(0,2)^2|$. The simplest way to see that is to show that $|(0,1)|=|(0,2)|$. –  Arturo Magidin Jun 20 '12 at 4:15
    
You seem to have a typo in your question: do you perhaps mean to ask about whether $|(0,1)|=|(0,2)|$? –  KReiser Jun 20 '12 at 4:22
    
No for the first part it is |(0,1)|=|(0,1)| And for 1 it is onto because onto means range f = B where B in that case would be {0,1} and range is also {0,1} It is confusing I know! thats why I am asking here. Thats all the information thats provided, and it asks whether the statement is true or false. So for the first one I think the statement is true because in order for (0,1)| = |(0,1)| to be a function it has to be bijective. –  user1411893 Jun 20 '12 at 4:24
    
Okay, that comment makes it even more confusing, because now we have $\{0,1\}$ thrown into the mix, as well. –  Cameron Buie Jun 20 '12 at 4:31
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You transcribed your first correction incorrectly; it compares the intervals (0,1) (that is, the open unit interval) and (0,1] (that is, the semi-closed unit interval); the latter is the former plus the point {1}. Notation is essential to get right for examples like these, particularly if you want us to be able to help you. –  Steven Stadnicki Jun 20 '12 at 6:59

2 Answers 2

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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Notational tip: if you have open intervals, use $\langle\cdot,\cdot\rangle$ for ordered pairs. –  Asaf Karagila Jun 20 '12 at 4:53
    
Excellent point. Fixed. –  Cameron Buie Jun 20 '12 at 6:03

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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I believe that the OP knows about the answer to the first question, as the OP also posted the question you linked to. –  Cameron Buie Jun 20 '12 at 16:02
    
@Cameron Heh. I missed that. Thanks. –  Rick Decker Jun 21 '12 at 17:18

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