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Show that $(a, b]$ and $[c, d)$ have the same cardinality, where $a, b, c,d$ are real numbers and $a <b, c<d.$

This is a well-known classical question in real analysis, if I am not wrong. The thing is, these problems are new to me and I am interested in learning some general approaches to establish such statements.

I could prove that any two closed intervals in $\Bbb R$ have the same cardinality, but till now, I haven't been able to solve this problem (above).

But, I have a feeling that all real intervals be it open intervals or closed intervals all have the same cardinality.

I hereby post an (incomplete) solution, I devised, for proving that $(a,b]$ and $(c,d]$ have the same cardinality:

We want to prove, that $(a, b]$ and $[c, d)$ have the same cardinality. We proceed by some linear transformations:

$(1)$ We transform, $(a,b]\to (0,b-a]$ using the transformation $t\to t - a$ then,

$(2)$ We consider the following shifting, i.e $(0,b -a]\to (0, 1]$ using the transformation $t\to\frac{t}{b-a}.$

$(3)$ Then we carry on, with some analogous reverse transformations, i.e $(0, 1] \to (0,d - c]$ using the transformation $t\to (d - c)t$ and

$(4)$ This last shifting, i.e $(0,d- c]\to (c, d]$ using the transformation $t\to t + c,$ let's us land to the desired place.

Now, as all of these transformations are bijections, so their composition is also so, which means, $(a, b]$ and $(c, d]$ does have the same cardinality. The bijection between them, as mentioned which is composition of these transformations in an ordered way, is precisely, $t\to \frac{d-c}{b-a}(t-a)+c$.

Now, I try to write the (above ) proof formally. This is as:

We consider a mapping $f:(a,b]\to (c,d]$ defined via, $$f(x)=\frac{d-c}{b-a}(x-a)+c.$$ Now, we must prove, that the function has it's range in $(c,d].$

If, $c\gt \frac{d-c}{b-a}(x-a)+c\implies a\lt x,$which is true. Similarly, if $\frac{d-c}{b-a}(x-a)+c\leq d\implies x\leq b$, which again is true, and these results let's us conclude, that $$c\lt \frac{d-c}{b-a}(x-a)+c\leq d.$$

Now, we claim: $f$ is well-defined, and a bijective mapping.

If we can prove, that $f$ is a bijection, we don't have to worry about any thing else. So, our aim is proving this as a bijection.

First, we try proving, that $f$ is an injection. If $x,y\in (a,b]$ such that $$f(x)=f(y)\implies \frac{d-c}{b-a}(x-a)+c=\frac{d-c}{b-a}(y-a)+c\implies \frac{d-c}{b-a}(x-a)=\frac{d-c}{b-a}(x-a)\implies x=y.$$ This proves, $f$ as an injection.

Our next ain, is to prove that $f$ is onto. We consider, $y\in(c,d],$ then if, $\exists x\in (a,b]$ such that $f(x)=y$ , it simply means that there exists an $x$ that satisfies $\frac{d-c}{b-a}(x-a)+c=y$ which further leads to the conclusion that $x=\frac{b-a}{d-c}(y-c)+a.$ This is true, if $\frac{b-a}{d-c}(y-c)+a$ satisfies the inequality, $a\lt x\leq b\implies $a\lt\frac{b-a}{d-c}(y-c)+a\leq b.$

Now, if this is true, then, for $a\lt \frac{b-a}{d-c}(y-c)+a\implies c\lt y,$ is true, , which is true. Also, if $\frac{b-a}{d-c}(y-c)+a\leq b$ is true, it means, $\frac{b-a}{d-c}(y-c)+a\lt b\implies y\lt d$, is true, which is true. So, $a\lt\frac{b-a}{d-c}(y-c)+a\leq b$ is true. It simply means, that $f$ is onto.

So, $f$ is a bijection. Hence, we get, $(a,b]$ and $(c,d]$ have the same cardinality.

Now, I want to know whether the thing I did, is a legit way or not?

If this stands, correct, then how do I proceed to prove that $(a, b]$ and $[c, d)$ have the same cardinality?

Thomas Finley
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    Can you find a linear map $f$ such that $f(a) = d$ and $f(b) = c$? – MPW Jun 13 '23 at 14:19
  • @MPW Forgive me, but I don't know, about the notion of a "linear map". – Thomas Finley Jun 13 '23 at 14:23
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    Take $f: (a, b] \mapsto [c, d)$ defined by $f (x) = \frac{c - d}{b - a} (x - a) + d$. Can you show that $f$ is a bijection? – K. Jiang Jun 13 '23 at 14:28
  • In this case, I mean a function whose graph is a line segment. It's formula would be something like $y=mx+b$, where $m$ and $b$ are numbers you need to determine in terms of $a,b,c,d$. You may want to look at this answer – MPW Jun 13 '23 at 14:28
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    Your general opinion is correct, but it's less straightforward to find a bijection between something like $[a,b]$ and $(g,h)$ (continuous maps won't work). I would say it's probably easier to find a surjection from $[a,b]$ to $(g,h)$ to show $|[a,b]| \ge |(g,h)|$ and an injection from $\mathbb{R}$ to $(g,h)$ to show $|(g,h)| \ge |\mathbb{R}| \ge |[a,b]|$. – user6247850 Jun 13 '23 at 14:59
  • @user6247850 If you consider posting a more formal solution, that'll be a big help for me! – Thomas Finley Jun 23 '23 at 15:26
  • @MPW Thanks! I could prove: $(a,b]$ and $(c,d]$ have the same cardinality. But how to prove, $(a,b]$ and $[c,d)$ has the same cardinality ? – Thomas Finley Jun 23 '23 at 15:27
  • Can you please quote exactly the definition of "have the same cardinality" that you are using? I'm pretty sure if you write it out, you should be able to answer your own question. – Stef Jun 23 '23 at 15:43
  • @Stef "Two sets $A$ and $B$ have the same cardinality iff there exists a bijective mapping between $A$ and $B$ ". I am using this definition, but forgive me, but I don't see how this simplify down the things in here? – Thomas Finley Jun 23 '23 at 15:46
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    @ThomasFinley If you want to prove $(a,b] \sim [c,d)$, define $f(b) = c$ and $f(x) = g(x)$ for $x\in (a,b)$ where $g$ is your favorite bijection from $(a,b)$ to $(c,d)$ – Andrew Jun 23 '23 at 16:20
  • By far the easiest way to approach problems like this is to use the following fact. If $A\sim B$ and $C$ is countable, then $A\sim (B\cup C)$ where $A,B$ are countable. And also the Cantor-Schroeder-Bernstein theorem: $A\sim B$ if you can find injections $f:A\to B$ and $g:B\to A$. Of course, that's a sledgehammer here, but the general way of thinking is useful and also applies to harder problems. – Andrew Jun 23 '23 at 16:23
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    Also, what user6247850 said is correct, but in light of my previous comment, is actually very straightforward to find an explicit bijection $f:[a,b]\to (g,h)$. Clearly we can just assume the unit interval, i.e. we may exhibit a bijection $f:[0,1]\to(0,1)$. Put $f(0) = 1/2$ and $f(1) = 1/3$ and $f(1/2) = 1/4$ and so on, i.e. $f(1/n) = 1/(n+2)$ where we define $1/0 = 0$. For $x\neq 1/n$, set $f(x) = x$. – Andrew Jun 23 '23 at 16:30

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