We all might be familiar with the beautiful method Cantor devised to prove that the cardinality of the set of real numbers is more than that of the set of natural numbers (Refer to: https://en.m.wikipedia.org/wiki/Cantor%27s_diagonal_argument). His method involved mapping each natural number to unique real number, no two natural numbers can map to same real number, and then showing that there are always real numbers left which are not assigned to any natural number.

But what can we say about the cardinalities of the sets $A$ and $B$ defined as $$A=\{x \ |\ x\in (0,1]\}$$ $$B=\{x \ |\ x\in [1,\infty)\}$$

Is the cardinality of $A,$ written as $|A|$ less than or equal to $|B|$?

I think that $|A|=|B|$ and I think so based on the following reasoning. Now I would use somewhat similar argument as Cantor did, in the sense that I will define one to one mapping of elements from set $A$ to set $B$.

Let us define a function, $f:[1,\infty)\rightarrow (0,1]$ defined by $$f(x)=\frac{1}{x}$$ It's easy to see that this is a one-one function. So we have mapped every real in $A$ to some unique real in $B$. And we have done this for all the elements of $A$ and $B$, as is clear from the fact that this is an invertible function.

So what's wrong in saying that this argument proves $|A|=|B|$?

Please express your thoughts on this.


  • 9
    $\begingroup$ Nothing is wrong with that argument. $\endgroup$
    – littleO
    May 21, 2022 at 1:56
  • 3
    $\begingroup$ Check out Are there fewer reals on $(0, 1)$ than on $(1,\infty)$? under the Related column on the right-hand side of this page. See also More numbers between $[0,1]$ or $[1,\infty)$? and one to one correspondance between points on the number axis and real numbers. $\endgroup$
    – dxiv
    May 21, 2022 at 2:05
  • 2
    $\begingroup$ There is nothing wrong with that argument. In fact, in two infinite sets A and B, as long as you can find two injections A->B and B-> A we say that they have the same cardinality, or i.e. if A\subset B and B\subset A, then |A|=|B|. $\endgroup$
    – LGu
    May 21, 2022 at 2:05
  • $\begingroup$ I understand your argument but does anyone ( including the OP ) care to comment on what the intuition is behind them having the same cardinality ? Clearly, most people not into math would probably think that the cardinality of the $[1,\infty)$ interval was greater than that of $(0,1]$. How can all of the numbers in the larger interval "fit" into the smaller interval ? $\endgroup$
    – mark leeds
    May 21, 2022 at 2:09
  • 4
    $\begingroup$ @mark, what you call intuition is based on our day-to-day experiences with finite sets. Infinite sets do not act in accord with these intuitions. What you need to do is to work with the mathematics of infinite sets for a few years, and you will develop a new set of intuitions, and at that point – but only at that point – the sort of result we're discussing here will become intuitively obvious to you. $\endgroup$ May 21, 2022 at 3:09

1 Answer 1


So what's wrong in saying that this argument proves $|A|=|B|$?

Absolutely nothing. If a one-to-one correspondence exists between two sets, then they have the same cardinality.

Infinite sets are just weird. For example, all of the following sets, even though intuitively having different sizes, have the same cardinality (denoted $\aleph_0$):

  • The integers $\mathbb{Z}$.
  • Subsets of $\mathbb{Z}$
    • The set of natural numbers $\mathbb{N}$.
    • The set of prime numbers.
    • The set of odd numbers.
    • The set of powers of 2.
    • The set of numbers that have a Collatz sequence that reaches 1, whether the conjecture turns out to be true or false.
    • The set of integers that, when written in binary and interpreted as a UTF-8 string, contain the name of a character from a Shakespeare play.
  • Supersets of $\mathbb{Z}$:
    • The set of rational numbers $\mathbb{Q}$.
    • The set of Gaussian integers.
    • The set of constructible numbers.
    • The set of computable numbers, under any computation model in which programs and their possible inputs and outputs consist of a finite but unbounded sequence of symbols from a finite alphabet.

Question: Can you think of a set of numbers that has more elements than $\mathbb{Z}$, but fewer than $\mathbb{R}$?

  • $\begingroup$ Thank you @Dan. I got it. $\endgroup$ May 21, 2022 at 4:46
  • $\begingroup$ may be this set $\mathbb{Z}\cup \{0.5\}$? $\endgroup$ May 21, 2022 at 8:17
  • 2
    $\begingroup$ It's a trick question, Shivam – don't fall for it! $\endgroup$ May 21, 2022 at 12:45
  • $\begingroup$ I see. Thanks. @Gerry $\endgroup$ May 22, 2022 at 6:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .