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It is my understanding that the cardinality of a set is the number of elements in a set. If I have a set $A = (0,1)$ and a set $B = [0,1]$, then how is it possible that $|A| = |B|$ if $A$ is missing two elements which are elements of B, namely 0 and 1?

The book I'm reading says it suffices to find two functions $f$ and $g$ such that $f: A \to B$ and $g : B \to A$. However, the example functions they give are $f(x)=x$ and $g(x) = \frac{x+1}{3}$. How can $f(x)$ map $(0,1) \to [0,1]$ if $0 \not\in (0,1)$?

jjagmath
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mars_plastic
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    If I had to guess, your book actually says it suffices to find two injective functions? In particular $f$ doesn't need to map something to $0$ – Brian Moehring Nov 21 '21 at 04:48
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    If you read carefully you'll probably find that $f$ and $g$ are required to be injective, not any two functions. The idea is that if $f:A\to B$ is injective, then $|A| \le |B|$ (which should be intuitive if you understand what injective means) – jjagmath Nov 21 '21 at 04:50
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    Also, the fact $A$ is missing $0$ and $1$ is not important. Actually, the bijection $x \mapsto 2x$ shows that $[0,1]$ and $[0,2]$ have the same cardinality, even if $[0,1]$ "is missing" an infinity of elements of $[0,2]$, namely all the numbers in $(1,2]$ – jjagmath Nov 21 '21 at 04:55
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    Does this answer your question? Bijection between an open and a closed interval – Peter O. Nov 21 '21 at 05:02
  • On the other hand, "it suffices" in this case means "By the Cantor-Schröder-Bernstein Theorem," which is routine but non-trivial. It's invaluable when a bijection is harder to find, but in this case we can find a simple bijection if that's what you'd prefer. – Brian Moehring Nov 21 '21 at 05:02
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    Does this answer your question? [How to define a bijection between $(0,1)$ and $(0,1]$?](https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1) – MJD Nov 21 '21 at 05:03
  • A crucial difference between finite and infinite sets is that the latter has the property that you can add/remove finitely many points to/from it without changing its cardinality. That is what separates the finite from the infinite. Delving deeper, you can add/remove countably-infinitely many points to/from an uncountable set without changing its cardinality. It might be counterintuitive at first to know that any interval in $\Bbb R$ has the same cardinality as that of $\Bbb R$ (an explicit bijection can be made by scaling and composing with a function like $\tan$) – Prasun Biswas Nov 21 '21 at 05:17
  • Hilbert's hotel might be an interesting read for you. – Prasun Biswas Nov 21 '21 at 05:18
  • @BrianMoehring Yes, the book states I can use that theorem. From what I understand of the theorem, if I can prove that $|A| \leq |B|$ and $|B| \leq |A|$, then |A|=|B|. I can see how $f(x)=x$ is injective and not surjective from $(0,1) \to [0,1]$, so $|(0,1)| \leq |[0,1]|$. What I don't understand is how to make an injective function that maps $[0,1] \to (0,1)$ – mars_plastic Nov 21 '21 at 05:27
  • $A=(0, 1) $ and $B=[0, 1]$

    Let, $S={1/n : n\in \mathbb{N}}$

    $S\subset B$

    Define a map, $f:B \to A $ by

    $ f(x) = \left{ \begin{array}{ll} \frac{1}{2} & x= 0 \ \frac{1}{3} & x= 1 \ \frac{1}{n+2} & x= \frac{1}{n} ({n>1})\ x & x \in B\setminus S\ \end{array} \right. $

    Now, study the map.

     – Sourav Ghosh Nov 21 '21 at 06:33
  • Interestingly, even though "two elements" is a lot more significant in the case of the countably infinite sets ${3,,4,,5,,6,,7,,\ldots }$ and ${1,,2,,3,,4,,5,\ldots }$ (because these sets are much smaller than $(0,1)$ and $[0,1]$ in the sense of cardinality), it's much easier to explicitly define a one-to-one correspondence between these two countably infinite sets. – Dave L. Renfro Nov 21 '21 at 08:24

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