Philosopher19

It makes no sense for one quantity of 10 to be bigger than another quantity of 10. 10 is one quantity. Similarly, it makes no sense for one quantity o...

This is where I disagree. I don't believe Cantor's diagonal argument shows anything. Infinity is one cardinality/size, it makes no sense for one infin...

Agree with all of the above. But you can't map one infinity to another with one being bigger than another because there isn't more than one. Infinity ...

It could be that arguing with someone is what it takes for them to be sincere to a truly perfect existence/being (or God/Goodness/Truth). It could be ...

I believe it makes perfect sense to say set x is only a member of itself in its own set, and I tired to prove it to you, and we even narrowed it down ...

I understand/agree. Let's focus on 3. There is a difference between: 3) In B, A is a set with 1 member, and that member is A/itself. (Correct) 3a) A i...

Forgot to say thank you for the detailed reply. Thanks.

So I think I understand what you mean by RANGE, but I feel any further conversation will not be fruitful. I feel like I've already said all that I sho...

Let me see if I've understood what you mean by RANGE. v = any set z = any set other than the set of all sets V = the set of all v Z = the set of all z...

That is not what I'm saying. Nor do I see what I'm saying as amounting to that. L does not have n members in LL. L is one item in LL. And of all the i...

L is a member of itself in L is true. L is a member of itself is true (but it is logically implied that L is a member of itself in L). It makes sense ...

If I ask the question where does L list itself, the answer is in L. If I ask where else does L list itself, the answer is nowhere else. This shows tha...

There is no difference. I asked the question to highlight a point about the property of "being a member of oneself". Every member of LL (including L) ...

I'm not asking how many members does x or y have. So I don't see how your example is relevant to what I asked. My questions were reasonable/relevant/m...

Ok

Here is what you said: See the parts I underlined? Here they are for more clarity: In B, A is not a member of anything in B, it is a member of B

You literally said two things that contradict each other: In B, A is not a member of anything. In B, A is a member of B. If in B A is not a member of ...

I believe you used your notation to avoid/miss the semantical point I was trying to make. Here are the answers to the questions I asked: Does L list i...

I believe this has nothing to do with what I said. I believe the exact answer to "Does L list itself in L?" is yes. I believe the conversation won't p...

So on the one hand you say: On the other hand you say: Do you see how you have contradicted yourself? And then you say Yes, proper attention has not b...

L is listed in both L and LL. In L, it has the property P of being a member of itself/L. As in the fact that L is in L is what instantiates the proper...

The property P is instantiated based on what set the item x is in. 1 and 2 were solid meaningful questions that highlight precisely this point. L is l...

Hi Michael In case you are interested in discussing this further: L = The list of all lists LL = The list of all lists that list themselves 1) In whic...

I don't think the process of continuing forever amounts to anything infinite. I see infinity as the reason for the process of continuing forever as be...

I watched a YouTube video amongst other things. Didn't read a book on it. Had a look at wiki and Stanford but maybe or maybe not with massive amounts ...

I believe I understand the matter well enough. I don't feel it sincere to Truth/Goodness to read/research any more on it than I already have. The link...

Already responded to it at least twice. Won't repeat myself.

I'm saying whether something is a member of itself or not is determined by whether it is in its own set or not. This matter of pure reason/definition ...

Fine, I will more or less repeat myself but this is probably the last time. p) In A, A is a member of itself/A. q) In U, A is not a member of itself/A...

It seems evident to me that you've not been paying attention to what I've been saying. I've already given more than one relevant reply to this. I'm no...

I have been saying to you that x is not a member of A if x is a member of x. I have not been saying 1. 1 is blatantly contradictory. So it seems to me...

I believe I addressed this point both in my reply to your non-math example and in your N and R example and in my other replies. No point doing in doin...

Am I the one that's confused? So it's not the case that in A it's a member of one thing and in B it's a member of another thing? So it's not the case ...

Maybe the following will help. I don't know.

I have not disagreed with scenario 2. I have said that in B, A is not a member of itself precisely because it is a member B (as opposed to itself), an...

Yet you seem to think that in B, A is both a member of A and B (which is contradictory)

I never said A can't be both a member of A and B. I said, in A, A is a member of A/itself, and in B, A is a member of B/other-than-itself.

I addressed a similar point in my reply to your non-math example. Every natural number can be both a member of N and R. I am not denying this. But wha...

And in B (as opposed to in both A and B), A is not a member of itself because it is a member of B (and not A). This is basic.

And now you have strayed from clear language. What I have given you is clear should you choose to pay attention to it: I asked you: to which you said:...

Look at what you're saying: In B, A is both a member of A and B. "In B" does not equal to "in both A and B". Do you see your contradiction? You have t...

But in B, A is a member of B. In B, A is not a member of both A and B. So once again, in B, is A a member of itself or not a member of itself?

Yes. But what you're not responding to is the following: In the case of A = {A}. A is a member of itself. In the case of B = {A, 0}, is A a member of ...

I believe you're not paying attention to what I'm saying. I'm not saying a set can't be a member of more than one set. I am saying: See again my repli...

So at least you're engaging me in clear meaningful language. I will respond in kind. You are not a set. You can never be a member of yourself. But a s...

I understand Russell's paradox. Here is what I say in response:

Evidently, in A, A is a member of itself. Evidently, in B, A is not a member of itself because A is a member of B. So you have not shown that 1 and 2 ...

My proof was here: As if "the rejection of the set of all sets is by definition contradictory" is not proof enough. Or as if "a set is not a member of...

I tried to look at your reply, but it is unclear to me as to what it's doing. I believe it deliberately strays from what is clear simple language to t...

That's because speaking in absolute terms, only the set of all sets is a member of itself. In short, if we were to focus on absolutely all sets, then ...