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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: https://i.stack.imgur.com/2MhxE.png

hmakholm left over Monica
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user1411893
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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 04:15
  • 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 04:22
  • 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 04:24
  • 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 04:31
  • When you write, "It is bijective", what doesn "It" refer to? Being bijective (or not) is a property of functions, and you haven't given one. – Gerry Myerson Jun 20 '12 at 04:42
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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 06:59
  • You also copied the second question incorrectly. I've edited the question to fix these errors. – Rick Decker Jun 20 '12 at 13:35
  • BTW the first part of the question was also poster here. – Martin Sleziak Jun 20 '12 at 14:59

2 Answers2

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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.

Cameron Buie
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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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Rick Decker
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