# Does the interval $[0,1]$ contains more real numbers than $(0,2)$? [duplicate]

It is a common saying that there are more real numbers than integers, since you can not uniquely map integers to real numbers - even after using up all integers, there will still be at least one real number left that you haven't mapped to any integer.

So I applied the same argument to the two intervals in question. One obvious way to map number from $[0,1]$ to $[0,2]$ is $\times 2$ - $0$ is still $0$, $0.25$ becomes $0.5$, and $1$ becomes $2$. Good enough.

However, if we change the second interval to an open interval, namely $(0,2)$, then the number $0$ and $1$ in the original mapping is now the "remaining" number, while all numbers in $(0,2)$ have already been "used". Does that mean that $[0,1]$ have more real numbers than $(0,2)$? (even though $(0,2)$ seems to be bigger than $[0,1]$…)

Does my argument even make sense? I did some Googling but didn't found anything useful.

## marked as duplicate by Nate Eldredge, Asaf Karagila♦, Ross Millikan, Mikhail Katz, Lord Shark the UnknownJun 29 '17 at 15:08

• It depends on what you mean by "more". The concept gets tricky with infinite sets. See math.stackexchange.com/questions/1754313/… to get started. – Nate Eldredge Jun 29 '17 at 14:52
• No, you've given one example of a function that isn't a bijection between $[0,1]\to (0,2)$, but to actually show that the latter has a different cardinality, you need to show that any function isn't a bijection between the two, not just find an example of one that isn't. It turns out that you can't do this because there is, in fact, a bijection between the two, it just isn't as nice as "times by 2". – Zain Patel Jun 29 '17 at 14:58

The map $x \mapsto \dfrac{x+1}{2}$ is a bijection $[0,1] \to [\dfrac12, 1] \subseteq (0,2)$.

Therefore, $(0,2)$ has at least as many elements as $[0,1]$.

The map $x \mapsto \dfrac{x}{2}$ is a bijection $(0,2) \to (0,1) \subseteq [0,1]$.

Therefore, $[0,1]$ has at least as many elements as $(0,2)$.

Thus*, the two intervals have the same number of elements.

(*) To make this rigorous, you can invoke the Cantor–Schröder–Bernstein theorem.

No, $[0,1]$ clearly does not contain more real numbers that $(0,2)$.

Notice that $[0,1]=(0,1]\cup \{0\}$ and $(0,2)=(0,1]\cup (1,2)$.