…then perhaps not everything can be infinite, which is where this thread veered off some time ago, in defining everything as capable of achieving infinitude.
That bolded part is a non sequitur…
To remove a member from a group necessarily changed WHICH members belong to that group… that is the specific way in which it changes the group.
I asked you if it was true or false… and your response was to make an unsubstantiated assertion, that does not answer the question.
Disagree.
Agree.
But it also necessarily changes the number of members in that group. You disagree with this for a reason that is unknown to me.
Note that I understand that the word “remove” on its own does not imply such a change. Rather, it’s the word “remove” together with the term “the number of elements” that does so. The standard meaning of the word “remove” together with the standard meaning of the term “the number of elements” implies such a change. By changing the meaning of one of these terms, you can change the relation between them. For example, you can redefine the term “the number of elements” in such a way (without redefining the term “remove”) so that removing a thing from a group of things has no impact on the number of things in that group. But such would be a deviation from the standard definitions.
Note that I provided an example of how removing a thing from a group consisting of an infinite number of things necessarily decreases the number of things in that group.
The direct answer to your question would be that your statement is false because your prior statements imply otherwise.
You said that (A = {1, 2, 3, \dotso}) (the infinite set of positive whole numbers) and that (B) is the same infinite set after we remove (y) amount of members from it. Let’s remove only (1) element and let that element be (1). The result is (B = {2, 3, 4, \dotso}).
The claim you put forward is:
For every member (x) in set (A), there is a member (x + 1) in set (B).
Like so:
1 → 2
2 → 3
3 → 4
etc
You asked me if your claim is true or false.
It’s false.
It’s false because you SAID that you produced set (B) by making a copy of set (A) and removing one element from it. This logically implies that (A) has every element that (B) has plus one more which in turn means that (B) is greater than (A). To say otherwise is to contradict this claim.
There’s no one-to-one correspondence between the two sets.
The following is the TRUE relation between the two sets:
1 → (none)
2 → 2
3 → 3
etc
That’s NOT a one-to-one correspondence.
And there is no way in hell you can get rid of this “(none)” on the right side. You can only falsely believe that you can get rid of it.
Magnus,
If 1 implies none, then that sequence is totally removed!
Your confusion is that you are enumerating things you define as inenumerable!
To be perfectly honest, at this point, you’re looking silly.
What does “1 implies none” mean?
What I’m saying is that (1) is not paired with any member from the other set.
- What sequence?
- What does it mean that it is totally removed?
What do you mean I am enumerating things?
How did I define things as inenumerable?
Alright let’s do this!
1.) 1
2.) 2
3.) 3
4.) 4
Etc…
1.) 1
2.) -
3.) 3
4.) -
Etc…
You’re numbering (enumerating) the dashes!
You say they don’t correspond to the set!
Then all you’re left with is this:
1.) 1
3.) 3
5.) 5
7.) 7
Etc…
But! This can be enumerated in correspondence!
1.) 1.) 1
2.) 3.) 3
3.) 5.) 5
4.) 7.) 7
Etc…
Understand now?
Let’s take it step by step.
Let (A) be the set of all natural numbers.
(A = {1, 2, 3, \dotso})
Let (B) be a set that is identical to (A).
(B = {1, 2, 3, \dotso})
Because the two sets are identical, it follows that they have exactly the same number of elements.
Since the two sets are identical, it follows that we can pair every element in (A) with exactly one element in (B) and every element in (B) with exactly one element in (A).
Like so:
1 → 1
2 → 2
3 → 3
etc
On the left, we have the elements of (A).
On the right, we have the elements of (B).
Do you agree so far? Yes or no.
Now, let us remove one element from (B).
Let that element be (1).
The result is this:
1 → (none)
2 → 2
3 → 3
etc
“(none)” here indicates that (1) is not paired with an element from (B).
The reason it is not paired with an element from (B) is because we REMOVED the element it was previously paired with.
Do you agree so far? Yes or no.
Since EVERY element in (B) is already taken (i.e. already paired with an element from (A)), there is NO element from (B) to pair (1) with.
This means the two sets do not have the same number of elements.
It’s pretty clear you do not agree with this.
The question is WHY.
You need to get rid of “(none)” in order to prove that the two sets have exactly the same number of elements.
And there is no way to do this. The best you can do is hide it inside the ellipsis.
You can do this:
1 → 2
2 → (none)
3 → 3
etc
You can do it one more time:
1 → 2
2 → 3
3 → (none)
etc
And one more time:
1 → 2
2 → 3
3 → 4
etc
Now you no longer SEE it because you hid it inside the ellipsis.
It didn’t cease to exist.
Either that or you’re simply DENYING that it exists.
There’s no way in hell you can get rid of it.
You can merely move it around or deny it.
This “(none)” does NOT belong to (B).
It means “No element in (B)”.
“1 → (none)” means “No element in (B) is paired with (1)”.
And the fact that it’s not an element in (B) does not mean it does not exist.
It exists and it cannot be eliminated.
Magnus, here’s where we disagree (and I think I’m correct (obviously))
Let’s say you remove an element:
1.)
2.) 2
3.) 3
4.) 4
Etc… we just removed the first element here!
The problem is, we have TONS of symbols to represent to space!
When we decide that we have NO symbols to represent that space, it’s considered NOT enumerable!
So we just remove the 1.), making it…
2.) 2
3.) 3
4.) 4
Etc…
It’s your work to explain a symbol that’s not a symbol so we can keep:
1.)
Making it enumerable but at the same time, not enumerable! It’s a contradiction!
What does this mean?
It means that autocorrect on a cellphone sucks!
Your definition is here:
“Not paired with any member from the other set”
If it’s not possibly paired, it’s not on a list in the first place!
Amongst other things, 1.) (blank) equals either zero or null!
If it cannot possibly be paired! Then 1.) (blank) is a blank of 1.)
Understand?
You keep pairing blanks! You can’t do this!
1.)
2.) 2
3.) 3
4.) 4
Etc…
You cannot LIST an UNLISTABLE!!!
What list?
It does NOT equal “zero” and it does NOT equal “null”.
Most importantly, it is not a member of any of the two sets.
“1 → (none)” simply means that (1) is not paired with an element from the other set. That’s all it means.
Consider the following two finite sets:
(A = {1, 2, 3, 4})
(B = {1, 2, 3})
There’s no one-to-one correspondence between them.
Try it.
1 → 1
2 → 2
3 → 3
4 → (none)
Regardless of how you pair them, there will always be an unpaired element. In the above case, this element is (4) and the fact that it is not paired with an element from the other set is represented by it being paired with “(none)” which means “No element from this set”.
You certainly wouldn’t say the two finite sets are equal in size, wouldn’t you?
I am not “pairing blanks”, I am simply saying that an element from one set is not paired with any element from the other set. In other words, I am merely saying that an element is UNPAIRED. “1 → (none)” simply means that (1) is unpaired (i.e. that it is not paired with any element from the other set.) It does NOT mean that it is paired with some element called “(none)”.
Magnus,
Even in finite sets (which have nothing to do with infinite sets)
I’d state that your (none) equals zero or null, as you are enumerating it!!!
If it’s “not paired with an element from the other set”, it’s not enumerated!
In order for that premise to be false, there must necessarily be at least one member x in set A such that there is not a member x+1 in set B
Which member would that be?
IF the premise was: for every member x in set A there is a member x in set B… THEN you would have proven me wrong.
But that was never the premise… this is what you might call a straw man.
This whole bijection business is nonsense for infinite sets.
Even if you matched element 1 of infinite Set A with any element of any position along infinite Set B, and maintained that same (or any different) interval for element 2 in Set A’s correspondence with Set B, and so on infinitely, you’d never run out of matches, regardless of which way around you set the sets.
So which is bigger than the other if it holds either way around?
This whole pretense of “well let’s make some arbitrary rule to make it seem like you couldn’t do this” is just that: arbitrary.
Infinite is infinite.
It lacks finitude, it’s arbitrarily bigger than any real number - there is no specificity to “that which is not finite/defined”.
The boolean “NOT” is binary - either there is quantity (which can be anything finite) or there isn’t. The latter cannot be anything finite without contradicting itself. A quantity of infinities is a contraction.
So let’s stop with this bullshit of bijection and one-to-one correspondence with infinite sets.
It’s so obviously contradictory.
It is possible to say what you’re saying without contradicting SOME of the previous statements.
However, it is NOT possible to do so without contradicting ALL of the previous statements.
This is a very subtle point and I am not sure I can communicate it well enough.
The mistake that you’re making is that you think that in order for your statement to be false it must contradict A SPECIFIC SUBSET of prior statements. In reality, it is enough for it to contradict ANY of the prior statements. (You are CONSTANTLY TURNING AWAY from those statements that your statement ACTUALLY contradicts, instead CHOOSING to focus on those it does not contradict. I think this phenomenon is known as confirmation bias.)
With that in mind, the answer to your question would be:
I agree that your statement DOES NOT contradict SOME of the prior statements. Your statement does not imply that there are elements in (B) for which we previously stated that they do not belong to (B). (2, 3, 4, \dotso) are all members of (B). This does not mean that your statement is true, it merely means that it is better than a statement such as “For every (x) in (A), there is (x - 1) in (B)” since that statement implies that (0) is a member of (B) which contradicts our earlier claim that (0) is not a member of (B).
You are stubbornly ignoring the fact that some of your prior statements logically imply that there is no one-to-one correspondence between the two sets.
You started with (A = {1, 2, 3, \dotso}) and (B = {1, 2, 3, \dotso}).
You paired the members like so:
1 → 1
2 → 2
3 → 3
etc
You removed (1) from (B) to get the following result:
1 → (none)
2 → 2
3 → 3
etc
The gap that is “(none)” cannot be eradicated for the simple reason that all other elements in (B) are already paired (i.e. no other elements in (B) are unpaired.).
You are asking whether the statement “For every (x) in (A), there is (x + 1) in (B)” is true. It is not. This is because there is an element (x) in (A) that is paired with “(none)” (which means it’s unpaired) instead of being paired with (x + 1).
Let’s do some step-by-step logic.
- Let’s pair every element of (A = {1, 2, 3, \dotso}) with an unpaired element from (B = {1, 2, 3, \dotso}).
1 → 1
2 → 2
3 → 3
etc
- Let’s remove (1) from (B).
1 → (none)
2 → 2
3 → 3
etc
- Let’s rename every (x) in (A) to (x + 1).
1 + 1 → (none)
2 + 1 → 2
3 + 1 → 3
etc
- Let’s make a mistake. Let’s substitute (1 + 1) with (2), (2 + 1) with (3) and so on.
2 → (none)
3 → 2
4 → 3
etc
This is a mistake because it leads to a logical contradiction. The above suggests that the two sets are BOTH equal in size and not equal in size.
In set theory, if two sets have exactly the same elements, they are said to be identical. And if two sets are identical, they are necessarily equal in size.
From the above, we can see that the two sets have exactly the same elements, so they are necessarily equal in size. However, at the same time, the presence of “(none)” suggests that they are necessarily unequal in size.
What this means is that we are not free to substitute (1 + 1) with (2). They are, in actuality, two distinct elements (they are NOT one and the same element.) Indeed, if go back to step 3, we can see that we renamed (2) to (2 + 1), not to (1 + 1).
It is precisely this sort of sophistry that creates the illusion that the two sets are equal in size. This unconscious process of renaming (or relabeling) of elements is something I spoke of pages ago (back when Gib participated in this discussion.) It’s a trick, nothing more, but it can be difficult to spot.
I have ignored no such thing… I have demonstrated beyond reasonable doubt that infinite sets do not behave the same as finite sets when members are removed.
You are incoherently insisting that we should accept as an AXIOM (by definition) that to remove a member from a group necessarily reduced the groups membership count.
I have shown you such an axiom would result in a contradiction with infinite sets… IOW it is tantamount to saying there can be no such thing as an infinite set.
Take your infinite queue of boys and girls holding hands… but let’s make it boys and their clone holding hands.
Boy1 ↔ Clone1
Boy2 ↔ Clone2
Boy3 ↔ Clone3
If we remove Clone1, then boy1 is left with no hand to hold. But if every single Clone in that infinite queue took one step forward (AT THE SAME TIME), then every boy would once again have a hand to hold.
In a finite set that would not be true because AT THE END of the queue there would be a lonely boy with no clone. But in an infinite queue there would be no lonely boy no matter how long you look for one… because there is no end.
You have stated that this premise is false because it contradicts the LOGICAL implications of “removing” a member according to your adopted axiom and I agree…
Because it results in a contradiction, is exactly why I think your axiom is stupid…
I’ll let you have the last word, if you want.
I am insisting that according to the standard definition of the word “remove” and the standard definition of the term “the number of things in a group”, to remove a thing from a group of things necessarily means to decrease the number of things in that group.
If my argument is wrong, then at least one of the following must be true:
- I got the standard definition of the word “remove” wrong
- I got the standard definition of the term “the number of things in a group” wrong
- I got the logic wrong (i.e. the conclusion does not logically follow from my definitional premises)
Which one is the case? Why? And how can we settle it?
That’s not true.
Boy1 ↔ Clone1
Boy2 ↔ Clone2
Boy3 ↔ Clone3
…
This means the two sets have the same number of elements. More importantly, it means that every BOY and every CLONE is paired with an element from the other set. There are NO unpaired boys and there are NO unpaired clones.
So, when you remove Clone1 to get this:
Boy1 ↔ (none)
Boy2 ↔ Clone2
Boy3 ↔ Clone3
…
There is NO unpaired clone to pair with Boy1.
Sure, you can move your clones around. They can move at the same time or they can move one at a time. It does not matter because either way you can’t get rid of “(none)”, which is to say, there will ALWAYS be an unpaired boy. All you can do is change who among the boys is not paired with one of the clones.
Are you denying that in order to re-establish one-to-one correspondence between the two sets, there must be an unpaired clone? Because if you’re denying this, you’re basically saying that we can pair Boy1 with a paired clone. The problem with this is that the result won’t be a one-to-one correspondence.
It’s very simple. You just have to stick to logic instead of trying to defend popular non-sense.
There will be a lonely boy at some point in the queue. It can be literally anywhere.
Boy1 ↔ (none)
Boy2 ↔ Clone2
Boy3 ↔ Clone3
…
Here, Boy1 is lonely.
Boy1 ↔ Clone2
Boy2 ↔ (none)
Boy3 ↔ Clone3
…
Here, Boy2 is lonely.
And so on.
The standard definitions become “stupid” the moment you realize you made a mistake that you do not want to accept. So instead of accepting that you made a mistake, you decide to change your definitional premises and declare that the standard definitions are “stupid”. The Fox and the Grapes.
If I say that all men are mortal and that Socrates is a man, it logically follows that Socrates is mortal.
The fact that I can declare (and yes, I can declare such a thing) and/or imagine (and yes, I can imagine such a thing) that Socrates is immortal does not mean that Socrates is indeed immortal. It proves nothing other than my inability to stick to logic.
My premises FORCE ME to adopt the position that Socrates is mortal. I am NOT free to adopt a different position. And the fact that I am able to adopt a different position does not make that position true, just as it does not make it consistent with my premises, just as it does not make my premises wrong.
In the same exact way, the fact that you can declare and/or imagine that (A = {1, 2, 3, \dotso}) and (B = {2, 3, 4, \dotso}) have the same number of elements proves absolutely nothing other than your inability to stick to logic.
You can declare the two sets are equal in size by relating them in the following way:
1 → 2
2 → 3
3 → 4
…
But you can ALSO declare that (B) is greater than (A) by relating them in the following way:
- → 2
1 → 3 - → 4
2 → 5 - → 6
…
This makes (A) two times smaller than (B).
The problem with BOTH declarations is that they CONTRADICT our earlier statements. That’s why they are wrong. None of them is right MERELY because we can make such a declaration.
Magnus,
You’re going to be so pissed at me
I’m going to state it again!
What does: —> symbolize to you?