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Is religion dying?

dybmh

דניאל יוסף בן מאיר הירש
do you agree with the definition?

OK. If you agree to the defintion I posted above, I'm quite sure it's checkmate in 8-10 moves. Maybe less. I haven't typed it out. But I think I have this completely resolved. {} is NOT a subset of any set A, and I can prove it. And that would invalidate any vacuous truth rendering it false.
 
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an anarchist

Your local loco.
I'm quite sure it's checkmate is 8 to 10 moves
God, I love this site

1683076904879.png
 

dybmh

דניאל יוסף בן מאיר הירש
OK, even better.

Let A = {} and B={1,2,3}.

What is their intersection?

CIao

- viole

Love it! They don't intersect, they are disjointed. They have nothing in common. { 1,2,3 } =/= { 1,2,3, {} }. That statement right there. That's checkmate in 1 move. {} is NOT the subset of any set.

Screenshot_20230503_063644.jpg

Screenshot_20230503_064435.jpg
 

dybmh

דניאל יוסף בן מאיר הירש
now, before I destroy whatever you wrote, as an alleged defeater of empty sets being subsets

Are you starting to see it? I have this. And there's nothing you can do to stop it.
 

viole

Ontological Naturalist
Premium Member
Love it! They don't intersect, they are disjointed. They have nothing in common. { 1,2,3 } =/= { 1,2,3, {} }. That statement right there. That's checkmate in 1 move. {} is NOT the subset of any set.

View attachment 76243

View attachment 76245
Are you joking? That is the whole point.

Of course they do not intersect. And what is the result of the intersection of two sets that have no element in common, and are therefore disjoint?

Spoiler: it is written in wikipedia too. Actually, it is written below the picture you posted lol

Let’s have a look:

In mathematics, two sets are said to be disjoint sets if they have no element in common. Equivalently, two disjoint sets are sets whose intersection is the empty set

Exactly my point. And you lose, according to your preferred defintion of subsets.



Ciao

- viole
 
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dybmh

דניאל יוסף בן מאיר הירש
Are you joking? That is the whole point.

Of course they do not intersect. And what is the result of the intersection of two sets that have no element in common, and are therefore disjoint?

Spoiler: it is written in wikipedia too. Actually, it is written below the picture you posted lol

Ciao

- viole

Good, we agree on the facts. You simply don't understand what they mean.

what is the result of the intersection of two sets that have no element in common, and are therefore disjoint?

The intersection of { 1,2,3 } and { 4,5,6 } = {}. That means by definition, the definitions and facts you have agreed to, {} is not a subset of either { 1,2,3 } or { 4,5,6 }.

Therefore {} is not a subset of any set. You have two examples which contradict the rule. And I can come up with infinite similar examples.

The miscomprehension is coming from confusing {} with { {} }. { {} } can be evaluated. {} cannot. Anytime you see {} on the left side of an operation, the author should be writing it as { {} }, but is leaving that out in error.

For example:

The intersection of {} and { 1,2,3 } cannot be evaluated.
The intersection of { {} } and { 1,2,3 } = {}.
{ {} } =/= {}.
 
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viole

Ontological Naturalist
Premium Member
Good, we agree on the facts. You simply don't understand what they mean.



The intersection of { 1,2,3 } and { 4,5,6 } = {}. That means by definition, the definitions and facts you have agreed to, {} is not a subset of either { 1,2,3 } or { 4,5,6 }.

Therefore {} is not a subset of any set. You have two examples which contradict the rule. And I can come up with infinite similar examples.

The miscomprehension is coming from confusing {} with { {} }. { {} } can be evaluated. {} cannot. Anytime you see {} on the left side of an operation, they author intends for it to be { {} }, but is leaving that out in error.

For example:

The intersection of {} and { 1,2,3 } cannot be evaluated.
The intersection of { {} } and { 1,2,3 } = {}. But { {} } =/= {}.
Of course it can be evaluated. All intersections of
sets give a set. As I showed you from the article. You sre just making things up, again, and persist in your state of denial. Despite the entire world disagreeing with you.

It is, a fact, that the intersection of the empty set with another set is the empty set. Always. Evidence is everywhere. Do you wanna see it? You can choose the source of internet addressing set intersection. Aren’t I magnanimous?

But, for fun, let’s use the second definition you proposed.

Again, let A={} and B={1,2,3}

What is their union?

Ciao

- viole
 
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dybmh

דניאל יוסף בן מאיר הירש
Of course it can be evaluated. All intersections beteen sets give a set. As I showed you from the article. You sre just making things up, again, and persist in your state of denial. Despite the entire world disagreeing with you.

No, I am using the definitions we have agreed on. I have not made anything up.

All intersections beteen sets give a set.

Disjointed sets do not intersect. So this is irrelevant. Talking about airplanes are equally irrelevant.


It is, a fact, that the intersection of the empty set with another set is the empty set. Always.

Which means the empty set is not a subset of any set. The empty set does not obtain all properties. I'm thinking you have made this up on your own.

Evidence is everywhere. Do you wanna see it?

Yes please. Perhaps one of those cute pictures you were talking about. Evidence, not claims would be very much appreciated. Otherwise you're preaching a religion.

You can choose the source of internet addressing set intersection. Aren’t I magnanimous?

OK, here is a picture showing an empty set.

Screenshot_20230503_121612.jpg

Set X = { 1,3,5,{} }
Set Y = { 2,4,6,{} }

The intersection of X and Y = { {} }. You can see this because the emptiness is bounded.

Set X =/= { 1,3,5 }. If so, the entire circle would be full, with no blank space. The same is true for set Y.

But, for fun, let’s use the second definition you proposed.

Again, let A={} and B={1,2,3}

What is their union?

It cannot be evaluated. {} is not defined in this context.

If the intersection of { 1 } and { 2 } = {} = { not 1, not 2, everything else }
Then the union of {} and { 1,2,3 } = { 3, and everything but 1 and 2 }

If the intersection of { 2 } and { 3 } = {} = { not 2, not 3, and everything else }
Then the union of {} and { 1,2,3 } = { 1, and everything but 2 and 3 }

However, { {} } is well defined.
The union of { {} } and { 1,2,3 } = { 1,2,3,{} }.

And... please note. The definition provided by wikipedia included 3 conditions, 3 properties which define a subset. All three need to be saticfied inorder to be considered a subset and to distinguish between a superset.
 
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dybmh

דניאל יוסף בן מאיר הירש
All the Jews I know are atheists.

Ciao

- viole

If you don't know any Jews then the set of Jews you know is completely divergent, disjointed, from the set of Atheists. Two disjointed sets have no correspondence, nothing in common. The word "are" makes the statement false.

The Jews you know = { {} }
Atheists = { Atheists }

The intersection of { {} } and { Atheists } = {}
{ {} } are { Atheists } is false.
{ {} } are { Atheists, {} } is true.
{ Atheists, {} } = "Atheists or not"

If you don't know any Jews, and you are honest, you should have said:

"All the Jews I know are Atheist or not."
 
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dybmh

דניאל יוסף בן מאיר הירש
Let’s have a look:

In mathematics, two sets are said to be disjoint sets if they have no element in common. Equivalently, two disjoint sets are sets whose intersection is the empty set

Yet again, you omitted out the part that proves you wrong.

It says: "Two sets A and B are disjoint if and only if their intersection is the empty set. It follows from this definition that every set is disjoint from the empty set, and that the empty set is the only set that is disjoint from itself.[5]

You have been saying that the empty set is a subset of every set. That is obviously contradicted here.
Further, if the empty set is disjointed with itself. That means {} =/= {} by defintion. Two sets are equal if and only if each set is a subset of the other. And that proves that {} cannot be evaluated.

In mathematics, two sets are said to be disjoint sets if they have no element in common. Equivalently, two disjoint sets are sets whose intersection is the empty set

Exactly my point. And you lose, according to your preferred defintion of subsets.

No. You have claimed that an empty set is a subset of all sets. What you posted above is the OPPOSITE of that. You said that it was so simple and easy to prove. You said it was taught to young children, and is included on the first 1 or 2 pages of a child's math book. So, go ahead. Prove it using the definition that we have agreed on. It should be simple, right? Just remember { 1,2,3 } =/= { 1,2,3,{} }.
 
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viole

Ontological Naturalist
Premium Member
Yet again, you omitted out the part that proves you wrong.

It says: "Two sets A and B are disjoint if and only if their intersection is the empty set. It follows from this definition that every set is disjoint from the empty set, and that the empty set is the only set that is disjoint from itself.[5]

You have been saying that the empty set is a subset of every set. That is obviously contradicted here.
Further, if the empty set is disjointed with itself. That means {} =/= {} by defintion. Two sets are equal if and only if each set is a subset of the other. And that proves that {} cannot be evaluated.



No. You have claimed that an empty set is a subset of all sets. What you posted above is the OPPOSITE of that. You said that it was so simple and easy to prove. You said it was taught to young children, and is included on the first 1 or 2 pages of a child's math book. So, go ahead. Prove it using the definition that we have agreed on. It should be simple, right? Just remember { 1,2,3 } =/= { 1,2,3,{} }.


There is no contradiction. Do you really believe that millions of mathematicians, logicians, including the greatest minds of the last two centuries, educators, students, etc. could have missed such a huge and obvious contradiction at the root of the fundaments of their discipline? I am sure you must have asked yourself how it can be that all articles, courses, books, etc. about the argument, agree with me, and none whatsoever agrees with you.

The source of your confusion is obvious. You rely too much, and erroneously, on the geometric visualization of sets as bubbles, as you can find in Venn and Euler diagrams. You must find it impossible to visualize a set that is contained in another set and, at the same time, is disjoint, external, to it. After all, a bubble cannot be inside and outside another bubble at the same time, right?

Well, in case of the empty set, this is the case. And it is the bubble analogy that fails, not the properties of the empty set. And it fails because the empty set should not be represented as a bubble. After all, if I have two non-empty sets which are disjoint, and therefore have the empty set as intersection, there is no third ball representing that intersection in any Venn diagram known to woman.

Therefore, best to not think of the empty set as a bubble. It will just mislead you.

You should simply rely on the definitions, like a robot. That is really the safest for you, and will put you back in line with the rest of the educated world.

So, let's take intersection and union as examples, and think only according to the definitions:

Q: What is the intersection of sets A and B?
A: A set
Q: What does that set contain?
A: All the elements, if any, that are common in A and B
Q: Can the result be an empty set?
A: Sure. if the two sets have no elements in common
Q: What if A is the empty set, and B is arbitrary?
A: Nothing. The definition still applies
Q: Will, in that case, the resulting set have elements?
A: Nope. Since if it had even one, it would be common with A, which is empty. And that would be absurd
Q: So, the result is an empty set, too. As in the case of not empty sets with no elements in common?
A: Of course

Ergo: --> {} intersection B = {} for all sets B

Q: What is the union of sets A and B?
A: A set
Q: What does that set contain?
A: All elements, if any, that are either in A. or in B. Or in both
Q: Can A be empty and B arbitrary?
A: Of course. The definition still applies
Q: Will the resulting set contain element of A?
A: Of course not, if A is empty and has, therefore, no elements
Q: So, it will contain only the elements of B?
A: Yes
Q: So, it will be equal to B?
A: Of course

Ergo: --> {} union B = B for all sets B

Any of those two obvious results entail immediately, according to the definition of subset you posted, that {} is a subset of B, for any B.

Now, this is basic stuff, as it is taught to kids as young as 10. You can, in fact, find the same results in Empty set - Academic Kids

  • For any set A, the empty set is a subset of A:∀A: {} ⊆ A
  • For any set A, the union of A with the empty set is A:∀A: A ∪ {} = A
  • For any set A, the intersection of A with the empty set is the empty set:∀A: A ∩ {} = {}


So, it is your call. Either you admit that you are incapable to understand what little kids are assumed to be able to grasp, or you ponder about it a little longer.

Possibly making a bit of research on your old school books, assuming you have ever been educated on this.

Ciao

- viole
 

Zwing

Active Member
Ergo: --> {} intersection B = {} for all sets B
I’ve been reading along with this. I now see what you mean, and have just now understood the fault in my thinking. It is simply this: I was premising that {}={0}, which is so dumb! There is no intersection of {1,2,3} and {0}; there is of {1,2,3} and {}, because {} is an element of every possible set. In other words (or rather, in words), the fact of a thing or things presumes nihility. Does this sound right?
 
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viole

Ontological Naturalist
Premium Member
I’ve been reading along with this. I now see what you mean, and have just now understood the fault in my thinking. It is simply this: I was premising that {}={0}, which is so dumb! There is no intersection of {1,2,3} and {0}; there is of {1,2,3} and {}, because {} is an element of every possible set. In other words (or rather, in words), the fact of a thing or things presumes nihility. Does this sound right?
Almost. It is a common to confuse the zero set with the empty set (aka null set). The zero set {0} contains one element, namely zero, which usually serves the purpose of neutrality to sums in some algebraic system, while the empty set {} contains no element whatsover. So, they are different objects. One has one element, the other has zero elements.

and it is not true that {1,2,3} does no have intersection with {0}. Intersections are always possible. And they always give sets. In this case, since the two sets have no element in common, the intersection is the empty set {}, too.

Ciao

- viole
 

mikkel_the_dane

My own religion

And now some sociology on the logical and illogical. Both can be observed for the Thomas theorem:
If men define situations as real, they are real in their consequences.

In everyday words, if I am illogical and act on that, it can be observed.
So the set for the everyday world includes the illogical and not just the logical.
 
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