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Let $x,y,a,b$ be real numbers. For $x and $a, show the cardinality of $[a,b]$ equals the cardinality of $[x,y]$.

I did the above problem, by defining a linear function as a bijection between the intervals. The next part of the problem states,

Extend your result (from above) to open and half-open intervals.

I'm just not sure what exactly I'm supposed to do next, the question seems vague. Am I supposed to show $(a,b)\sim (x,y)$? $[a,b]\sim (x,y)$? $[a,b)\sim (x,y]$?

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    I think so. They do all have the same cardinality anyway, same as the closed interval I mean.2012-10-01
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    I’m not sure myself how that’s intended to be interpreted, but I suspect that it just wants you to show that any two non-empty open intervals have the same cardinality and that any two non-empty half-open intervals have the same cardinality, regardless of their direction. All of that can be done using the method that you’ve already discovered. Finding a bijection between intervals of two different types is a bit harder.2012-10-01
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    You may need to ask the problem-poser. The only non-trivial problem is if the number of closed ends does not match. Then one can use a little trick related to Hilbert's Infinite Hotel.2012-10-01
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    Okay, thank you. And can I use the same function that I used for the closed intervals? f(t)=[(y-x)(t-a)]/(b-a) +x2012-10-01
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    @Alti: If the end-points, or lack of them, match, then sure. A simple variant takes care of $[a,b)$ and $(x,y]$. But if numbers of endpoints don't match, like $[a,b)$ and $(x,y)$, you will need a way to suck $a$ into the interior, but that will displace a point in $$(x,y)$, and so on.2012-10-01

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