# Is a number preceding infinity, finite?

1. Oct 22, 2011

### King

Hi,

I'm not sure if this is the right section, but I'm talking about numbers :).

The questions is as written in the title: Is a number preceding infinity, finite?

2. Oct 22, 2011

### pwsnafu

tl,dr version: no "number" (whatever that means) precedes infinity (which infinity?).

Long version: When you use the word "precede" it means there is a unique element which is less than it, but still larger than all others contenders. So on the integers, 3 precedes 4, because 3 < 4 but 3 > 1 and 2 . 10001 precedes 10002. And so on.

But on the rational numbers, nothing precedes, say, 2.
Why? Suppose there was, call it x. Then $x < \frac{2+x}{2} < 2$. So x can't precede it.

Similarly, take the smallest infinity $\aleph_0$. If n is a natural number that precedes it, when what about n+1?

What you can do, it look at larger infinities. So $\aleph_0$ precedes $\aleph_1$ which precedes $\aleph_2$ and so on.

3. Oct 22, 2011

### King

Why is this? If x=0 the inequality makes sense.

I see it is as anything preceding infinity is finite. The reason is because inifinity is defined as a never ending value. Nothing can be greater than a never ending value, but I know plenty of values that are not never ending, and hence must be less than infinity. Of course infinity has some kind of strange philosophical aspect attached to it, and it sounds as though its separated from every other number.

4. Oct 23, 2011

### pwsnafu

The inequality is true for any rational number less than two. That's the point.
If x=0, ie, we argue that that x=0 is the predecessor to 2, then the inequality gives 0<1<2, contradicting the definition of "preceding".
No matter what value of x we take, we can find a rational between the two.
"preceding" is not the same thing as "is less than". You are thinking of the latter.
To qualify for the former, there must be nothing in between the two values.

Edit: On second thoughts, scratch this. I'm thinking too much into it. Let's just use "less than" and be done with it.

That is not the definition of infinity. Importantly, you are using the word "infinity" in the singular. There are many infinities, some larger than others. $\aleph_0 < \aleph_1 < \aleph_2 < \ldots$

We have two FAQs on this: https://www.physicsforums.com/showthread.php?t=510966" [Broken]
Note that every real number has a never ending decimal representation. If we used your definition then every real number is infinite.
Just because a property is true for some numbers does not mean it is true for every number.

Last edited by a moderator: May 5, 2017
5. Oct 23, 2011

### lavinia

what other possibility would there be?

6. Oct 23, 2011

### pwsnafu

Infinite. For example the infinity $\aleph_0$ is strictly less than the infinity $\aleph_1$, but none the less $\aleph_0$ is still an infinity.

7. Oct 23, 2011

### phinds

So you believe pwsnafu is wrong, yes? why is that? What math do you use to say you are right and he is wrong?

8. Oct 23, 2011

### lavinia

the posts here discuss the real numbers not the ordinals.

9. Oct 23, 2011

### lavinia

I didn't say anybody was wrong. There needs to be some ordering of the extended real numbers before you can discuss what number precedes another number. this post seems to be discussing te extended reals rather than ordinals.

Maybe I don't understand your post. Explain.

10. Oct 23, 2011

### phinds

The problem w/ answering the OPs question is, as explained by pwsnafu, the term "number preceding infinity". The point, as I see it, is that infinity minus 10 is STILL infinity, so I claim that "a number preceding infinity" is infinite, as does pwsnafu. You, on the other hand, seem to be saying that it is finite. I don't see how you can say that, and I don't see any confusion about my contending that you have said pwsnafu and I are wrong. We contend the answer is "infinite" and you contend that the answer is "finite". We can't both be right.

I'm not a mathematician, and I don't understand why you say "There needs to be some ordering of the extended real numbers before you can discuss what number precedes another number" in terms of THIS discussion, but since I DON'T understand it, maybe I'm missing something.

11. Oct 23, 2011

### lavinia

Infinity minus 10 has no meaning. When one says that a number precedes another number it must preceded it in some ordering.

Technically infinity is not a number unless you are talking about the ordinals. But in this post the numbers seem to be the reals plus infinity. But then why not minus infinity as well? And is minus infinity less than plus infinity? Are the finite numbers greater than minus infinity? Is minus infinity equal to plus infinity?

There is no arithmetic for the numbers plus infinity.

12. Oct 23, 2011

### pwsnafu

Firstly, we have many ways of treating infinity as a number: ordinals, extended reals, real projectives, surreals, and the Riemann sphere to name a few. All of these define operations with it, so it is an arithmetic. And by the axiom of choice, there always is a well ordering. Whether we can write that ordering down is a another matter.

The problem with the OP was that he didn't specify which infinity he was talking about. If he wanted to talk about the extended reals so be it. But he must learn there are other possibilities.

Now, if we are to restrict the discussion to the extended reals, the answer is still no. The http://en.wikipedia.org/wiki/Extended_real_number_line" [Broken] are the real numbers adjoined with positive and negative infinity. We define $-\infty < x < \infty$ for all $x \in \mathbb{R}$. So negative infinity is an infinite that is less than positive infinity.

Last edited by a moderator: May 5, 2017
13. Oct 23, 2011

### King

Regarding the definition of preceding:
1. Come before (something) in time.
2. Come before in order or position.
It doesn't have to be one position before. However, it should be safe to say that a number preceding another (where the numbers are ordered from lowest value to greatest value) has a lesser value. Hence, is the same as saying that it is less than.

Let me be a little bit more specific about my question. Given a set of real numbers, the largest possible number would be the (base - 1) recurring - using decimal, this would be 9 recurring. Because 9 is recurring indefinitely, surely one can state that it is infinite? I suppose using this definition the same could be said about 1 recurring, but one would argue that this value is still less than 9 recurring. I suppose this answers my question...1 recurring is infinite and is less than 9 recurring, hence a number less than infinity is not necessarily finite...

Does anyone agree with what I just said?

Hmm...perhaps I should be talking about natural numbers to avoid the negative numbers which do not apply in my scenario. What do the books of mathematics state about infinity in a set of natural numbers?

14. Oct 23, 2011

### ramsey2879

The links you cited don't give any information on infinity. I agree that infinity $\aleph_0$ and $\aleph_1$ are both infinity, but it seems to me to be wrong to say that one infinity is less than another since infinity is not a number and it makes no sense to talk about infinity plus or minus a numerical value. Where do you get that one positive infinity is less than another positive infinity?
As I see it $\aleph_0$ and $\aleph_1$ are the same value.

Last edited: Oct 23, 2011
15. Oct 23, 2011

### ramsey2879

If you are talking of never ending decimal representatations with a fixed decimal point, you are talking of a finite limit which is not an infinite value, but a finite value.

16. Oct 23, 2011

### pwsnafu

To King: in mathematics the term "successor" really does mean "one position after", for example, in Peano Axioms the successor function takes a number and returns the next number. "Predecessor" would mean something something similar, and in fact, it does in my field of work. So please, don't use the term.

That is not the definition of infinity people on this thread have been using. But if that is the definition you want to use, you are correct. Every real number has an infinity long decimal expansion. But that is not a useful definition

The definition of $\aleph_1$ is the cardinality of the set of ordinal numbers. This is uncountable hence is strictly greater than $\aleph_0$. Applying the axiom of choice we can conclude there is nothing in between. http://en.wikipedia.org/wiki/Aleph_number" [Broken]

Last edited by a moderator: May 5, 2017
17. Oct 23, 2011

### phinds

Well, the thing is, math operations don't really CARE what you think and they do not follow your "thought".

Aleph null and Aleph one are NOT the same value. I suggust you read up on them.

18. Oct 23, 2011

### ramsey2879

Fine. Now we both know what we are talking about. If the cardinality of the countable natural numbers is $\aleph_0$ then $\aleph_0$ plus or minus 10 is also countable; you just count 10 more or 10 less. Therefore, $\aleph_0$ + 10 is the same value as $\aleph_0$. That is what I was trying to say.

Last edited by a moderator: May 5, 2017
19. Oct 23, 2011

### King

I have concluded that I am working with integers. Can infinity be a member of the set of integers? As you may be able to tell, I am no mathematician, hence I have not been trained to understand infinity as some of you have. If infinity is not viewed as a number, then what is it? How else would you describe a number of unbounded size?

I'm truly trying to get my head around it, so please bare with me!

20. Oct 23, 2011

### lavinia

So what is the arithmetic of the extended reals?

What is the arithmetic of the Riemann sphere?