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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.

maowtm
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  • It depends on what you mean by "more". The concept gets tricky with infinite sets. See https://math.stackexchange.com/questions/1754313/are-we-allowed-to-compare-infinities to get started. – Nate Eldredge Jun 29 '17 at 14:52
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    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

2 Answers2

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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.

lhf
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    See https://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval for an explicit bijection. – lhf Jun 29 '17 at 15:05
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No, $[0,1]$ does not contain more real numbers that $(0,2)$.

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

Asinomás
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