Questions about the union of these sets?

[Dr.Peterson]

Just out of curiosity, does the union of [0, 0.3], [0, 0.33], [0, 0.333], [0, 0.3333] ... equal [0, 0.333 ...] or put another way as [0, 1/3]?

Think! How does the same reasoning you've seen apply here? Is 1/3 a member of this union?

I just wanted to know if that poster was correct in conversation I had. The poster wrote that the union is in a fully closed interval as [0, 0.999...]. But it seems unanimous here that it is half open as you have.

No, I don't think he said that (though you clearly didn't show us the part you really have in mind). He said this,

There is no contradiction - it's just that the operation "take the interval [1, x]" is not continuous: the limit of the intervals is not the interval you get from the limit.

(which may agree with what we've said), and this,

It may (or may not) help your intuition there to see it from the other side. If we take the intervals [0.1,1], [0.01, 1], [0.001, 1], [0.0001, 1], and so on, and overlay all those intervals, the result does not include 0. We never find 0 in a single interval. So overlaying all those intervals gives (0,1]. However, I'd guess you agree that 0.000... is just zero!

which as I read it says what we've said. Otherwise, he's talking about limits, not intervals.

What is not clear is what you had said previously that he disagreed with! Can you show us the rest?

 

[Mates]

Think! How does the same reasoning you've seen apply here? Is 1/3 a member of this union?

Which reasoning are you referring to? I think 1/3 is a member, but I am not sure. I asked because I am not sure.

No, I don't think he said that (though you clearly didn't show us the part you really have in mind). He said this,

(which may agree with what we've said), and this,

which as I read it says what we've said. Otherwise, he's talking about limits, not intervals.

What is not clear is what you had said previously that he disagreed with! Can you show us the rest?

I could have swore I saw the poster write that the union does equal [0, 0.999 ...], but I don't see it there anymore. In fact, that was the whole reason I poster here. I wanted to double check to see if it exists or not. I had my doubts because of the obvious contradictions that would arise.

 

[blamocur]

Just out of curiosity, does the union of [0, 0.3], [0, 0.33], [0, 0.333], [0, 0.3333] ... equal [0, 0.333 ...] or put another way as [0, 1/3]?

It's the same as before, the union is equal to semi-open interval [imath][0, 1/3)[/imath], not closed interval [imath][0, 1/3][/imath].

 

[blamocur]

I think so, but I am not sure. I asked because I am not sure.

If 1/3 belonged to the union then it would have to belong to at least one of the member sets of that union -- which one would that be?

 

[Mates]

If 1/3 belonged to the union then it would have to belong to at least one of the member sets of that union -- which one would that be?

Thanks for the answer, but pka put a formula with the pi notation showing that an infinite number of products somehow does not include 0.999 ... I don't know how this is makes sense.

 

[blamocur]

Thanks for the answer, but pka put a formula with the pi notation showing that an infinite number of products somehow does not include 0.999 ... I don't know how this is makes sense.

I don't know which formula you mean, or how it is related to my post.

 

[Mates]

I don't know which formula you mean, or how it is related to my post.

Pka put " [imath]\displaystyle\bigcup\limits_{k = 1}^{\infty} {\left[ {0,1 - \frac{1}{{{{10}^k}}}} \right]}=\left[0,0.\overline{\,9~}\right) [/imath] Note that this is a half-open interval (one not included). "

I don't understand how we can use an infinite k, but not have infinite 9's (behind 0) included in the interval.

 

[Dr.Peterson]

Thanks for the answer, but pka put a formula with the pi notation showing that an infinite number of products somehow does not include 0.999 ... I don't know how this is makes sense.

Apparently you mean this:

We use to spend a great deal of time on this topic in our foundations course.
For example: [imath]\displaystyle\bigcup\limits_{k = 1}^5 {\left[ {0,1 - \frac{1}{{{{10}^k}}}} \right]}=[0,.9]\cup[0,.99]\cup[0,.999]\cup[0,.9999]\cup[0,.99999]=[0,.99999] [/imath]
So that we get [imath]\displaystyle\bigcup\limits_{k = 1}^{\infty} {\left[ {0,1 - \frac{1}{{{{10}^k}}}} \right]}=\left[0,0.\overline{\,9~}\right) [/imath] Note that this is a half-open interval (one not included).

Those aren't pi, representing products; they are unions. The first says the union of a finite number of closed intervals is a closed interval; the second says the union of an infinite number of closed intervals is a half-open interval that does not include 0.999... (that is, 1).

Now you've quoted it:

Pka put " [imath]\displaystyle\bigcup\limits_{k = 1}^{\infty} {\left[ {0,1 - \frac{1}{{{{10}^k}}}} \right]}=\left[0,0.\overline{\,9~}\right) [/imath] Note that this is a half-open interval (one not included). "

I don't understand how we can use an infinite k, but not have infinite 9's (behind 0) included in the interval.

But this is exactly what we've been saying! I thought you agreed that it doesn't include 1??

I just wanted to know if that poster was correct in conversation I had. The poster wrote that the union is in a fully closed interval as [0, 0.999...]. But it seems unanimous here that it is half open as you have.

It is a union of infinitely many sets, none of which contain 1. They all have a number with a finite number of digits as their end point; 0.999... is not one of those. And as you know, 0.999... = 1.

 

[Mates]

Apparently you mean this:

Those aren't pi, representing products; they are unions. The first says the union of a finite number of closed intervals is a closed interval; the second says the union of an infinite number of closed intervals is a half-open interval that does not include 0.999... (that is, 1).

Now you've quoted it:

But this is exactly what we've been saying! I thought you agreed that it doesn't include 1??

It is a union of infinitely many sets, none of which contain 1. They all have a number with a finite number of digits as their end point; 0.999... is not one of those. And as you know, 0.999... = 1.

Right, I got mixed up. (It's strange what I think I am seeing versus what I am seeing). Anyways, my issue with pka's formula is how we can use an infinite k and not have an infinite 9's (behind 0) in a interval.

 

[Dr.Peterson]

Anyways, my issue with pka's formula is how we can use an infinite k and not have an infinite 9's (behind 0) in a interval.

Which I answered! Read it again:

It is a union of infinitely many sets, none of which contain 1. They all have a number with a finite number of digits as their end point; 0.999... is not one of those. And as you know, 0.999... = 1.

Yes, it takes time to get used to the oddities of infinity; you have to stop and think often. Don't rush through it.

The union of sets is the set of all elements that are in at least one of them. The number 0.999... = 1 is not in any of these, so it is not in the union, no matter how much it may feel like it should. The endpoints of all those intervals are 0.999...9, with a finite number of digits. There are infinitely many of them, but none of them is itself infinite.

In just the same way, there are infinitely many natural numbers, but none of them is itself infinite. Infinity is not a natural number; it is the "number" (cardinality) of them.

Don't answer me too quickly; take time to think through it.

 

[Mates]

Which I answered! Read it again:

Yes, it takes time to get used to the oddities of infinity; you have to stop and think often. Don't rush through it.

The union of sets is the set of all elements that are in at least one of them. The number 0.999... = 1 is not in any of these, so it is not in the union, no matter how much it may feel like it should. The endpoints of all those intervals are 0.999...9, with a finite number of digits. There are infinitely many of them, but none of them is itself infinite.

In just the same way, there are infinitely many natural numbers, but none of them is itself infinite. Infinity is not a natural number; it is the "number" (cardinality) of them.

Don't answer me too quickly; take time to think through it.

Thanks for helping with my confusion as well as my approach to learning. Answering too quickly has been a constant issue with me. I will think hard about what you said here.

 

[Agent Smith]

[imath]\{a\} \cup \{b\} = \{a, b\}[/imath] Ergo, [imath]\{0.9\} \cup \{0.09\} = \{0.9, 0.09\}[/imath]

Also [imath]n(\{0.9\} \cup \{0.09\}) = 2[/imath]

 

[Mates]

[imath]\{a\} \cup \{b\} = \{a, b\}[/imath] Ergo, [imath]\{0.9\} \cup \{0.09\} = \{0.9, 0.09\}[/imath]

Also [imath]n(\{0.9\} \cup \{0.09\}) = 2[/imath]

I don't understand what you are saying

 

[Dr.Peterson]

I don't understand what you are saying

They're saying that if, in your original question, [0.9] meant {0.9}. the set containing only this one number (which is a plausible interpretation), then the union would just be the set of those numbers you listed. The union of two of those sets is a set containing 2 numbers.

But of course, that is not what you meant by that, so you can ignore it.

On the other hand, if you genuinely don't understand the meaning of what they wrote (as opposed to its relevance to your question, which is lacking), then you need to study set notation ...

 

[Agent Smith]

I don't understand what you are saying

Same here and hence the comment.

This would've made sense to me.
Let [imath]n(A) = 0.9[/imath] and [imath]n(B) = 0.09[/imath]
[imath]n(A) + n(B) = 0.9 + 0.09 = 0.99[/imath]

 

[Mates]

Same here and hence the comment.

This would've made sense to me.
Let [imath]n(A) = 0.9[/imath] and [imath]n(B) = 0.09[/imath]
[imath]n(A) + n(B) = 0.9 + 0.09 = 0.99[/imath]

Interesting. Can we keep summing like that to infinity so that there are an infinite number of 9's behind the decimal?

 

[Agent Smith]

They're saying that if, in your original question, [0.9] meant {0.9}. the set containing only this one number (which is a plausible interpretation), then the union would just be the set of those numbers you listed. The union of two of those sets is a set containing 2 numbers.

But of course, that is not what you meant by that, so you can ignore it.

On the other hand, if you genuinely don't understand the meaning of what they wrote (as opposed to its relevance to your question, which is lacking), then you need to study set notation ...

@Mates

Gracias Dr. Peterson for clarifying the matter.

To be fair, the OP seems to have serendipitously (miracles do happen) discovered how addition is defined in set theory. Assuming, of course, he's NOT aware of the apposite ideas and simply had his wires crossed.

For 2 disjoint sets A and B with n(A) = a and n(B) = b, the addition a + b = n(A U B). Correct/Incorrect/Both/Neither?

 

[Dr.Peterson]

@Mates

To be fair, the OP seems to have serendipitously (miracles do happen) discovered how addition is defined in set theory. Assuming, of course, he's NOT aware of the apposite ideas and simply had his wires crossed.

For 2 disjoint sets A and B with n(A) = a and n(B) = b, the addition a + b = n(A U B). Correct/Incorrect/Both/Neither?

No, he just didn't say what he meant. You're reading into it things that are not there. Did you read the rest of the discussion? He's talking about intervals, and used the wrong notation.

Same here and hence the comment.

This would've made sense to me.
Let [imath]n(A) = 0.9[/imath] and [imath]n(B) = 0.09[/imath]
[imath]n(A) + n(B) = 0.9 + 0.09 = 0.99[/imath]

How do you make sense of the cardinality of a set being a non-integer??

 

[Agent Smith]

[math]n(B) = 2[/math]No, he just didn't say what he meant. You're reading into it things that are not there. Did you read the rest of the discussion? He's talking about intervals, and used the wrong notation.


How do you make sense of the cardinality of a set being a non-integer??

Which is a superb question. Very continuum hypothesisish. A, B, and C are [math]3[/math] sets such that [math]n(A) = 1[/math] and [math]n(C) = 2[/math] and [math]n(A) < n(B) < n(C)[/math]

 

[Mates]

Which is a superb question. Very continuum hypothesisish. A, B, and C are [math]3[/math] sets such that [math]n(A) = 1[/math] and [math]n(C) = 2[/math] and [math]n(A) < n(B) < n(C)[/math]

I am not sure if you saw my question (please excuse me if you did). I am interested to know if we can sum to infinity with the method you showed in post #35. Would that make the number 0.999 ...?