The Solution to Russell’s Paradox, and the Absurdity of More than One Infinity

The overlooked part of the paradox

If x, y, and z are sets that are not members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. I cannot write x = {x, y, z} because x is in x, which makes x a member of itself.

If x, y, and z are sets that are members of themselves, and I form a set of these three sets, to represent this, I can write something like: p = {x, y, z}. Consistency with the previous paragraph dictates that I cannot write x = {x, y, z} because x is in x, which makes x a member of itself twice (which is contradictory as nothing is a member of itself twice, or nothing is itself twice). 

You cannot have a set of all sets that are not members of themselves because it will result in at least one set not being included in the set. In other words, x will have to be included in x, but it can’t.

You cannot have a set of all sets that are members of themselves because it will result in at least one set being a member of itself twice. In other words, x will have to not be included in x, but it can’t.

The importance of reference with regards to determining whether something is a member of itself or not

Take L to be the list of all lists, and take L’ to be the list of all lists other than L. Take V to be the set of all sets, and take V’ to be the set of all sets other than V. 

L (the list of all lists) is a member of itself only as a list, not as a set (and not just any list. L is only a member of itself as L. L is obviously not a member of itself as L’). So L is not a set, nor is it L’. V’ is not V. In the context or reference point of of V, V’ should not be viewed as a member of itself. I will attempt to illustrate this further:

Call any set that is not a member of itself a -V. Call any set that is not the set of all sets a V’. Call any set that’s simply a set, a V (the V of all Vs = the set of all sets). 

Is the V of all -Vs a member of itself? I would say yes because the V of all -Vs contains all -Vs and it is a member of itself. However, set theorists may object to this and say “when we say the set of all sets that are not members of themselves, we mean to say a set that consists of all sets that are not members of themselves, and no other sets”. What such set theorists want, is contradictory. You cannot have a set of all sets that are not members of themselves that is itself not a member of itself. In other words, you cannot have a -V as a V of all -Vs. The -V of all -Vs, is not the same as the V of all -Vs. The former is necessarily contradictory whereas the latter is not necessarily contradictory. For some reason, this has led them to conclude that you cannot have a V of all Vs (which is absurd, and they know this, yet they accept it or settle for it).

No V, or V’, or -V, can encompass all -Vs and nothing more. But one V can encompass all -Vs and something more. The V of all Vs encompasses all -Vs as well as itself.

Do two Vs encompass all V’s? 

One V (which is a V’) encompasses all V’s and nothing more. The other V (which is not a V’) encompasses all V’s and something more. The latter is the set of all sets (the V of all Vs), the former is the V’ of all V’s (the not-the-set-of-all-sets set of all not-the-set-of-all-sets sets). Thus, only one V encompasses all Vs and nothing more (the V of all Vs). Only one V’ encompasses all V’s and nothing more (the V’ of all V’s).

The above shows that whilst there can be no -V that encompasses all -Vs, there is a V that encompasses all -Vs. Whilst there can be two Vs that encompass all –V’s, there can only be one V’ that encompasses all -V’s. -V and –V’ are semantically not the same (neither are V and V’). -V’ = any V’ that is not a member of itself. 

Interpreting Russell’s question

If Russell was suggesting that a V of all -Vs must encompass all -Vs and nothing more (which is the equivalent of demanding a -V of all -Vs), then we cannot have a V of all -Vs. Consistency would then have us accept that we cannot have a V of all V’s (because only the V’ of all V’s can encompass all V’s and nothing more. So whilst we can have a V’ of all V’s, we cannot have a V of all V’s, just as we could not have a V of all -Vs). Ultimately, this would mean we can only have a V of all Vs. The conclusion is the same: When the reference is V, only one V is a member of itself (the V of all Vs).

When the reference is V’, only one V’ is a member of itself (the V’ of all V’s). When the reference is -V, no -V is a member of itself. -V is only meaningful in the context of V (just as -V’ is only meaningful in the context of V’). It is in the definition of -V that it is not a member of itself precisely because it is a member of V. So how can there be a -V of all -Vs? Rejecting the V of all Vs is contradictory on all fronts. It is the last thing we should be doing.

Conclusion

Is the set of all sets that are not members of themselves a member of itself? The -V of all -Vs is contradictory, but the V of all -Vs is not contradictory, and it is a member of itself. If you reject the notion of a V of all -Vs, then as a set, only the V of all Vs is a member of itself, and it is the only set that encompasses all -Vs.

Thus, if we are to be absolute with our standards or reference in relation to sets, then only the V of all Vs is a member of itself as a V (which would mean no other set is a member of itself). If we lowered our standards, then many Vs can be interpreted as members of themselves (V’ was one such example, but then again V and V’ are not the same. It is wrong to say V’ is a member of itself when it is a member of the V of all Vs. As members of the V of all Vs, all V’s are -Vs). By definition, no V that is a -V, or any other V that takes a hyphen before it (like -V’) can ever be a member of itself in an absolute sense. More absolutely, no V other than the V of all Vs can be a member of itself as a set (V).

-V is only absolutely true in the context of the absolute V. Where you wrongly take V’ as your absolute, -V’ is absolutely true. The V of all Vs encompasses one more -V than V’. This is another possible justification for saying -V is only absolutely true in the context of the absolute V (the V of all Vs). -V’ is only absolute in the context of the V’ of all -V’s, because only the V’ of all -V’s encompasses all -V’s and nothing more. Though I would have preferred not to say that the V of all Vs encompasses one more -V than the V’ of all V’s (because there are no -Vs in the context of V’s, there are only -V’s in the context of V’s), I thought it appropriate to say it here to highlight a particular point of view regarding one’s approach to sets.

The universal set

Call absolutely any thing (number, shape, tree, human, dream, colour) an ‘existent’. Call the set of all existents, ‘Existence’. Note that I am not referring to how real something is/exists, just that it is an existent (a member of Existence). Numbers are numbers (which is the same as saying numbers exist as numbers in Existence). The alternative is to say numbers don’t exist in Existence, or that there is/exists no such thing as numbers, or that numbers are in non-Existence (like round squares and other absurdities).

All existents (including Existence) are a member of Existence (because they are all existents). Only Existence is a member of itself as an existent. Since no other existent is a member of itself as an existent, the set of all existents that are not members of themselves, is Existence (and it is a member of itself. It is self-existing or self-sustaining or self-contingent).

By definition, Existence has no beginning and no end. Rejecting this yields contradictions:

It is hypothetically possible to have more than one galaxy, planet, or universe, but it is impossible to have more than one “Existence”. By “Existence” I mean that which all things exist because of or as a result of. Without Existence, nothing would encompass or unify all things into one Existence. This would mean that it is possible for one set of existents to be in existent A, and another set of existents to be in existent B, such that no existent encompasses A and B. Since no existent encompasses A and B, this means that non-existence separates A from B. For non-existence to separate A from B, it would have to exist. It is contradictory/absurd (semantically inconsistent) to say non-existence separates A from B because non-existence does not exist for it to do this. Hence the necessary existence of Existence. Semantics exist in Existence, as do imaginary unicorns (I imagined one just now). How real something is in Existence, is another matter. In any case, if x exists, then it either belongs to Existence, or it is Existence.

Anything that is contradictory (semantically inconsistent), is wrong by definition. If we want to believe in a finite Existence, we might as well believe in a triangle with only two sides.

Existence is a meaning, so it is a member of the set of all meanings. But then again, Existence IS the set of all meanings because there is no other thing, existent, set, or meaning that existentially contains all meanings. The set of all ducks is not some existing animal or shape. The set of all ducks is Existence Itself (which is an existing meaning/set/existent/truth). In other words, all ducks (imaginary, dream, or otherwise) exist in Existence. An imaginary duck exists as an imaginary duck. Dreams and imaginary ducks may not exist/be as real as us, but they are not non-existents. Since only Existence is truly infinite, Existence/Infinity is the set of all cardinalities.

By definition, lies and absurdities are not true of Existence, but they are existents (lies exist in Existence, what they describe does not). It is absurd to say there is no set of all sets (or true universal set) when there is only one Existence, and a complete/true infinity of existents.

Infinity versus semi-infinity

= The set of all numbers (if it’s a number, then it is included in this set)

= The set of all numbers except the number 19

It seems that both A and B encompass an endless number of numbers. I will attempt to show:

1) B is what I will call a semi-infinite set, whilst A is an infinite set.

2) Semi-infinites (or semi-infinite sets) come in various sizes, but there is only one infinity (so there aren’t infinities of various sizes, nor is there more than one infinite set). 

If you tell me “there is no end to the number of numbers that B encompasses”, and I ask you “does B encompass the number 19?”, you will say “no”. To which I will say “if there’s no end to the number of numbers that B encompasses, why doesn’t it encompass 19? Had you said ‘excluding 19, there is no end to the number of numbers that B encompasses’ I might have believed you”.

Whilst there absolutely/truly is no end to the number of numbers that A encompasses, there is an end to the number of numbers that B encompasses in an absolute sense. Having said that, the number of numbers that B encompasses is not finite in quantity (hence the term semi-infinite). Furthermore, B is one possible maximally large semi-infinite set of numbers (because it encompasses all numbers but one, and there are an endless number of semi-infinite sets that do this. A semi-infinite set that encompasses all numbers but two is smaller than the aforementioned semi-infinite set).

3 comments

  1. The Lemming · September 24

    Salam/Peace dear brother. I am so sorry if this article went over my head, i am technically somewhat philosophical illiterate but this is a argument against polytheism?

    Like

  2. Anonymous · September 25

    Peace brother

    It is an argument against there being more than one truly infinite being. So you could say it’s an argument against polytheism if you take the truly infinite being to be God.

    Liked by 1 person

    • The Lemming · 14 Days Ago

      Excellent, we need more articles refuting the polytheism, i have not seen much articles doing monotheistic arguments and showing polytheism does not make sense and why monotheism makes FULL LOGICAL SENSE.

      Like

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