I know the question sounds a little stupid, but yesterday laying on my bed this question came up in my mind ( I had watched the movie pi before it). We know that there surely are more real numbers between 0 and 2 than between 0 and 1, but still how can you prove something like that, like mathematically prove. You can't just take let

x = real numbers between 0 and 1 

and since

2 > 1, thus real numbers between 0 and 2 > x.

Because x is infinity. And adding, subtracting anything to x dosent make any sense, because it is infinite. How do you suppose we solve this.

  • 1
    $\begingroup$ The map $x\mapsto 2x$ establishes a $1:1$ correspondence between the intervals $[0,1]$ and $[0,2]$. That's what is meant by saying that there are "the same number of reals in the two intervals." $\endgroup$
    – lulu
    Apr 11, 2020 at 16:49
  • 3
    $\begingroup$ This depends on what you mean by "more". In set theoretic terms, there are exactly the same number. In measure theoretic terms (with the usual measure), the answer is essentially by definition. $\endgroup$ Apr 11, 2020 at 16:49
  • $\begingroup$ We can't...${}{}{}$ $\endgroup$
    – DonAntonio
    Apr 11, 2020 at 16:49
  • $\begingroup$ Does this answer your question? Explaining Infinite Sets and The Fault in Our Stars $\endgroup$ Apr 12, 2020 at 0:55
  • $\begingroup$ Also: Comparing numbers between (0 and 2) and (0 and 1). $\endgroup$ Apr 12, 2020 at 0:56

3 Answers 3


In fact, there aren't "more" (in the sense of cardinality) real numbers in $[0,2]$ than in $[0,1]$. There is exactly the "same amount of numbers " in the two sets. Nevertheless, $[0,1] $ is by sure a proper subset of $[0,2]$. You should become acquainted with the idea that this is not a contradiction. I suggest to start with the article of wikipedia: https://en.wikipedia.org/wiki/Cardinality


There are exactly same number of elements in both the intervals as there exists a bijection $f:[0,2]\rightarrow [0,1]$ defined by $f(x)=x/2$. For each number $x$ you choose from $[0,2]$, one can give you a unique number $x/2$ from $[0,1]$. For instance, $0\rightarrow 0, 1/2\rightarrow 1/4, 1/3\rightarrow 1/6,......,2\rightarrow 1$.


As answered by @diegog7, they are actually same in terms of their set cardinality. In fact any interval over real number line can be mapped to the entire real number line using one-one functions. For Example : (-1,1) can be mapped to R using one-one function : tan(πx/2) Similarly, any number on real number line can be mapped to interval (0,1) using the sigmoid function. Sigmoid function is defined as 1/(1+e^(-x)).

Using the result stated above, it is further claimed that any there are exactly equal number of reals between any two distinct reals.

Now, coming to your question about real numbers between A=[0,1] and B=[0,2] : I give a bijection f:A-->B as f= 2x. You can see that every element of A has been mapped to a unique element in B. Thus, proving that their cardinalities are actually equal and they had to be because they were individually equal to R.

Now, a reasonable doubt would be what if the function proposed was f=x (seems cardinality of A is less than that of B) or f=3x (seems cardinality of A is greater than that of B) . Well, I never actually intended to define cardinality that way for infinite sets. Greater than and less than are actually much stronger relations than (greater than or equal to) and (less than or equal to) in case of infinite sets. To prove strict inequalities like A greater than B you need to show that there exists no injection from A to B.

Actually it's better to say in the case of f=x that |A| ≤|B| and in f=2x that |A|≥|B|, both of which lead to the result that |A|=|B|.

Now proving this way might seem trivial to you but this is a very important result in the study of cardinalities of infinite sets and is popularly known as Schröeder-Bernstein theorem.

Link to Schröeder-Bernstein proof https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.whitman.edu/mathematics/higher_math_online/section04.09.html&ved=2ahUKEwj1wsKC5OHoAhXr6nMBHVYrDAQQFjAHegQIAhAB&usg=AOvVaw3bIHH8zE1se3z8H_gV8O4Z


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