/sci/ - Science & Math » Thread #11535909

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Anonymous Sun 05 Apr 19:29:11 2020 No.11535909 View Reply Original Report

Quoted By: >>11536034 >>11536084 >>11536117 >>11536324 >>11536378 >>11536395 >>11536465

Is there any such thing as a set with a finite number of numbers that does not have a largest number?

Anonymous

Anonymous Sun 05 Apr 2020 20:17:54 No.11536034 Report

Quoted By: >>11536542

>>11535909
Sets of complex numbers.

Anonymous

Anonymous Sun 05 Apr 2020 20:30:34 No.11536084 Report

Quoted By: >>11536089

>>11535909
https://en.m.wikipedia.org/wiki/Greatest_and_least_elements

Anonymous

Anonymous Sun 05 Apr 2020 20:32:00 No.11536089 Report

Quoted By: >>11536542

>>11536084
Thanks for googling for me. Where on that page does it answer my question, idiot?

Anonymous

Anonymous Sun 05 Apr 2020 20:38:55 No.11536117 Report

Quoted By: >>11536196 >>11536284

>>11535909
Ask that question in a different way and you'll have your answer.
Can any number or any thing that has inherent properties lose those properties for no reason? No. If you have a set of 10 1s, 1 is the largest number by definition of no other larger number being in there. If you have a red apple in your set and add in a green apple, the red apple didn't lose any properties aside from being the only one in the set which has nothing to do with your question.

Anonymous

Anonymous Sun 05 Apr 2020 21:09:02 No.11536196 Report

Quoted By:

>>11536117
>Ask that question in a different way and you'll have your answer
Thanks, dalai fucking lama

Anonymous

Anonymous Sun 05 Apr 2020 21:27:51 No.11536284 Report

Quoted By:

>>11536117
The post means

Is there any such thing as a set with a finite number of different numbers that does not have a largest number?

Anonymous

Anonymous Sun 05 Apr 2020 21:36:38 No.11536324 Report

Quoted By: >>11536343 >>11536357

>>11535909
Let $S$ be a finite set of cardinality $n$ . Let $s_0 \in S$ . Given that $s_0$ is not the largest element in the set there exists $s_1 \geq s_0$ .

Applying this recursively to $s_1$ to get $s_2$ we can find elements such that
$s_n \geq s_{n-1} \geq ... \geq s_1 \geq s_0$ .

But there are $n+1$ $s_i's$ . This means that there exist $i \neq j$ such that $s_i = s_j$ .

Now remove this $s_i$ from the set and apply the same procedure to the resulting set with n-1 elements. You will find that all elements are equal. This is a contradiction.

Anonymous

Anonymous Sun 05 Apr 2020 21:44:00 No.11536343 Report

Quoted By:

>>11536324
>Now remove this <math xmlns="http://www.w3.org/1998/Math/MathML">;
<msub>
<mi>s</mi>
<mi>i</mi>
</msub>
</math>
which?

Anonymous

Anonymous Sun 05 Apr 2020 21:47:15 No.11536357 Report

Quoted By:

>>11536324
>Now remove this $s_i$
which?

Anonymous

Anonymous Sun 05 Apr 2020 21:53:59 No.11536378 Report

Quoted By:

>>11535909
if by number you mean real number, than no

Anonymous

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Anonymous Sun 05 Apr 2020 21:58:42 No.11536395 Report

Quoted By: >>11536427

>>11535909

For goods notions of finiteness, there exists totally ordered sets which admit some finite subset without a maximal element.

Anonymous

Anonymous Sun 05 Apr 2020 22:06:13 No.11536427 Report

Quoted By: >>11536513

>>11536395
Next question:
Was/is goods an idiot?

Anonymous

Anonymous Sun 05 Apr 2020 22:14:01 No.11536465 Report

Quoted By: >>11536488

>>11535909
>Is there any such thing as a set with a finite number of numbers that does not have a largest number?
Yes, the empty set.

Anonymous

Anonymous Sun 05 Apr 2020 22:18:45 No.11536488 Report

Quoted By: >>11536548

>>11536465
One must question whether the empty set is a set with a finite number of numbers.
>0
still. Is it also a set of everything, having existed or not, in the same way i.e. being 0 of those things?

Anonymous

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Anonymous Sun 05 Apr 2020 22:29:59 No.11536513 Report

Quoted By: >>11536650

>>11536427

I know at least 17 definitions of "being a idiot" that are not equivalent in an arbitrary topoi.

Anonymous

Anonymous Sun 05 Apr 2020 22:41:06 No.11536542 Report

Quoted By: >>11536650

>>11536089
Why are you ignoring my answer?
>>11536034

Anonymous

Anonymous Sun 05 Apr 2020 22:43:45 No.11536548 Report

Quoted By: >>11536650

>>11536488
>One must question whether the empty set is a set with a finite number of numbers.
Yes. And then one must answer that question with "yes".

>Is it also a set of everything, having existed or not, in the same way i.e. being 0 of those things?
No.

Anonymous

Anonymous Sun 05 Apr 2020 23:12:19 No.11536650 Report

Quoted By: >>11537835

>>11536513
Was he any of them, idiot?
>>11536542
Your answer seems based on the notion that complex numbers don't have a greater or lesser value in relation to each other, which is a fallacy of incredulity until actually shown that they don't, i.e. what is or is not known of any element implies absolutely nothing about that element. It's otherwise a good and simple answer and thus far less interesting.
>>11536548
>No.
No with a falsely stern period isn't good enough for me. Elaborate.

Anonymous

Anonymous Mon 06 Apr 2020 02:39:01 No.11537351 Report

Quoted By: >>11537579

Yes. A set with several elements tied for largest. i.e. [1,2,3,17,17]

Anonymous

Anonymous Mon 06 Apr 2020 04:17:55 No.11537579 Report

Quoted By:

>>11537351
t. retard
repeated elements aren't allowed in a set only a multiset.

Anonymous

Anonymous Mon 06 Apr 2020 07:02:38 No.11537835 Report

Quoted By:

>>11536650
>Your answer seems based on the notion that complex numbers don't have a greater or lesser value in relation to each other
Yes.
>It's otherwise a good and simple answer and thus far less interesting.
Okay :-(

Anonymous

Anonymous Mon 06 Apr 2020 08:49:35 No.11538023 Report

Quoted By:

yes.
[1,i]

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