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If $A$ is the set of all real numbers in $(0,1)$ with no $5$ in their decimal representation, and $B$ is the set with no $34$ and no $76446$. Then the set $B$ is in some sense larger then $A$, how can I express this mathematically? How can I compare the relative infinities of these sets?

  • Is $A$ a set of open intervals?
  • Is there a bijective function $A \leftrightarrow B$?
  • Is there always a bijective function from a (potentiallly infinite) set of open intervals on the real number line to $(0,1)$?
  • What if some interval is closed?
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there is a bijection between A and B –  yoyo Apr 11 '11 at 20:38

3 Answers 3

up vote 2 down vote accepted

Edit: The answer to the questions above are that the two sets have the same cardinality. See the other answers for the answers to the specific questions you asked.

This answer shows how you can distinguish the sizes of the sets using a probability method.

Picking two random numbers independently from $(0,1)$, what is the probability that the entire interval between will be excluded from your set? This is only true if the two numbers "fail" at the same digits, so I think you get different probability measures this way, depending on your starting set. Essentially, you are excluding intervals from $[0,1]$ which add up to size $1$ in both your sets, but by excluding the squares of those intervals, you get a value less than $1$ and greater than $0$.

For example, in the case of your first set, if the two numbers you pick are in the interval $(0.5,0.6)$, then the entire interval between them is outside of your set. So the probability of picking a pair of numbers satisfying this condition is at least $0.01$, but cannot add up to $1$ because if $a\in (0.5,0.6)$ and $b\in(0.6,0.7)$ then there is always an element of your set between them.

I think this probability will measure these sets in the way you intend.

Edit: (From my comment below.) Okay, I don't like this answer, after all. For example, if you take the set $S_1$ to be the real numbers in (0,1) without a 5 or 6 in the decimal representation, you get a different value, from my method, than if you take the set $S_2$, the set with no 5 or 7 in the decimal representation. That strikes me as wrong - these two sets should not really be considered different in size, intuitively, but my computation comes up different for these two sets.

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Probability doesn't help since the probability of both sets is zero. –  Yuval Filmus Apr 11 '11 at 21:28
    
Not if you take the probability that I define in my post. Yes, the probability of a single point being in your set is zero, but the probability I define above is not zero and not one. –  Thomas Andrews Apr 11 '11 at 21:35
    
Think in terms of intervals removed for $(0,1)$ to get $A$, and then realize that if you remove the squares of those intervals fom $(0,1)\times (0,1)$ then you get a total interval size removed from the unit square is not $0$ or $1.$ –  Thomas Andrews Apr 11 '11 at 21:41
    
You're right, I didn't read past the first line. Nice idea! –  Yuval Filmus Apr 12 '11 at 3:00
    
It's not clear to me what this probability represents, and I've asked a question related to it without a satisfactory answer at: math.stackexchange.com/questions/32413/… . Is this measurement useful in any way outside distinguishing 'sizes' of sets like this - sets that are kind-of like the Cantor Set? –  Thomas Andrews Apr 13 '11 at 16:44

It appears you have mixed some things: You describe $A$ as a set of real numbers and then ask if it is a set of open intervals. Open intervals are subsets of the real numbers and therefore $A$ cannot be a set that contains open intervals (it contains real numbers). Maybe you are asking if $A$ is a union of open intervals? In that case the answer to whether $A$ is such a set is no because between any two real numbers there is a real number with $5$ in its representation. Likewise for $B$.

The cardinality of the two sets is the same as the cardinality of the continuum. To see this simply take into consideration the fact that the numbers that contain only $1$ and $0$ in their decimal representation belong in these sets and can be put into an one-to-one correspondence with every number in $(0,1)$ using the binary representation of the real numbers.

No, there is not an always a function from a potentially infinite set of open intervals to the open interval $(0,1)$. Any collection of disjoint open intervals is at most countable since each of them contains rational numbers. But any set that is a union of open intervals has the same number of elements as the open interval $(0,1)$ since the cardinality of any open interval is that of the continuum.

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It sounds like you are asking whether $A$ is a union of open intervals. The answer is no, as every interval contains (continuum many) points not in $A$. As Apostolos says, the cardinality of $A$ and $B$ is the same, and the definition of this is that there is a bijection between them. Similarly, as all of $\mathbb{R}$ can be put in bijection with $(0,1)$, and any interval contains continuum many points, any subset of $\mathbb{R}$ that contains an interval can be put in bijection with $(0,1)$. Whether an interval is closed makes no difference as long as you do not ask for continuous bijections. An explicit bijection $(0,1)\leftrightarrow (0,1]$ is $\frac{1}{2^n}\rightarrow \frac{1}{2^{n-1}}$ for $n \in \{1,2,3,dots\}$ and all other numbers stay the same.

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