Prop: Show that the intervals $(0, 1)$ and $(1, 3)$ have the same cardinality.
What I have tried: I showed that the intervals $[2, 4]$ and $[0, 5]$ have equal cardinality by creating a function $F(x) = \frac52x - 5$. I did this by getting the length of both intervals, however, I don't understand why this works. In my notes, it shows that the length of $[2, 4]$ is $2$ but if we are including both endpoints shouldn't it be $3$? Same goes for the interval $[0, 5]$ we should have a length of $6$, not $5$, but when using $5/2$ and shifting everything by $-5$ it seems to work just fine.
Trying to solve this current prop: Because it is an open interval we cant use $0$ and $1$ from what I understand so I don't quite understand what to use as the length of this interval.
I have tried reading other questions similar to this one however I am still stumped.
Thanks for the support!
You are right, the cardinality is "the number of elements within a set" (see here). For example, for the finite sets: $|\{1,2,3\}|=|\{5,6,7\}|=3$, because there are three elements in each set. Also, you can say the two sets have equal cardinality because there is a bijection from the first set to the second. Both of these sets are countable, because their cardinalities are finite. The term "denumerable" usually refers to "countably infinite" or $\aleph_0$ (the cardinality of the natural numbers), which does not apply here. The set of rational numbers is also denumerable (see here). See also this post for distinguishing the concepts.
The idea is the same for infinite sets. Note that $(0,1)$ and $(1,3)$ are uncountable, hence not denumerable. Yet, we can establish the equality of their cardinalities if we can find a bijection from one set to another. And we can, it is $f(x)=2x+1$, because every element of the first set can be injectively and surjectively matched with the elements of the second set. See the graph:
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How do we find the linear relation? Just find the line passing through the two points: $$\frac{y-1}{3-1}=\frac{x-0}{1-0} \Rightarrow y=2x+1.$$