Case 1
$B_1, B_2 \in \{D(z,\epsilon)\}$,
case (a) $B_1 \cap B_2=\phi$
case (b)$B_1 \cap B_2\neq \phi$
$x\in B_1 \cap B_2$.
Geometrically, I am able to see an element of $\{D(z,\epsilon)\}$,$B_3(say)$ such a way that $x\in B_3$ and $B_3\subset B_1 \cap B_2$. I took radius of $B_3$ as $d...
@LeakyNun Okay, yeah makes sense. So you won't get credit/not eligible for seminars etc? I really wanted to do some third/fourth year modules early, but read that regulations prohibited this. :\
have to study things that arent in our concentration frequently but we often get a lot of leeway in what those things are and tbh i think that's a good thing
I've studied/registered for 16 modules so far, and out of those 14 were compulsory; the other 2 were out of 4 possible choices in total - so really not much of a choice there either.
@Mathgeek: You should ideally give a formula for the radius in terms of the distance of $x$ from the centers of the balls, rather than saying distance to boundary. You can be explicit. You wrote nothing in case 4. Don't you also need to say every point is in some basis element? (Obviously, that's clear, but ...)
in the book the author makes a "false" quote, what if alot of history is just a fake, and then they come up with technologies to verify the age of a document, but to a mortal it could be as well magic.
I was surprised myself. They were carrying what seemed like 10 gallons jug of water. We got into this conversation about the benefits of water and I said that. They were very concerned.
@TedShifrin Haha, I usually truly flamboyant and incredible lies, usually I'm joking but I've a perpetual poker face so I let it take its course when I see that it wasn't received as such.
Leaky: I think one important consequence of ZF-empty or ZFC-Empty is that intersections can never be empty (because the empty set does not exist). This will have implications in topology, for example, we can no longer have T1 spaces
@LeakyNun But axiom of empty set asserts the existence of empty set, so nuking that (plus suitable modification to axiom of infinity) should nuke the empty set
@TedShifrin You mentioned that Apple computers are well built which I agree. But the Apple Superdrive which is an external DVD drive is not well built! I read so many bad reviews about it, how it cannot detect discs.
Is there a version of set theory that allows the existence of a set that does not admit the empty set as a member? I.e., reject the axiom $A\cup \emptyset = A$
We have some set X with a minimum and is of cardinality A. It has $A$ subsets of cardinality $B<A$, each has a mimimum, and each of these subsets have $A^2$ subsets of cardinality $C<B$ and so on the construction goes
Let's say X is a set with size A and with A subsets of size B<A and A subsets of size C<B<A. Does the next level is the level of B or C? And what if B,C are not comparable?
That is, let X be the set and Y be the disjoint union of the subsets of size A and Z be the disjoint union of subsets of size B. Then we have Z subset Y subset X
X has this tree like structure under the subset relation
No, let me give you an example: we have N the set of natural numbers. Then Nn=all of the subsets of size n. With this I am defining level as Nn is a level and the next level is Nm for m<n. With this my next level is not unique
@Secret for this we need linear order, because it is possible that there are disjoint maximal chains
Assuming choice, we can choose the next level. But when choice fail we need better rule, maybe if A is a set such that all of its subsets are aleph numbers in size it will work?
No, this is not good enough, we need to somehow choose the next level after limit aleph of limit ordinal
@Secret It is sufficient, but you need first to define the structure, rn the structure uses choice (in the "next level" part) if you successfully find a structure such that the conditions holds it is d-infinite
The dream said the tree is uncountable in height but it has not said anything more about what kind of uncountable cardinal it is and I disagree with the dream that it is an aleph
Suppose you have a finite group, and you've manually listed each pairing and assigned each one a particular value, ensuring that pairings where any two pairings where the order of one is the reverse of the other pairing have been assigned the same value. Is it still possible for that group to be non-abelian?
Now, I'm not sure if the hight will be aleph number, as it is possible that there exists a level with is d-finite union aleph number, and from there I have no idea how the tree will behave
Well, I can give a quick example: $\{I, A, A^{-1}\}$, where I define $AA^{-1}=A^{-1}A=I, AA=A^{-1}, A^{-1}A^{-1}=A, AI=IA=A, A^{-1}I=IA^{-1}=A^{-1},$ and $II=I$. This is a manual pairing group that, unless I am mistaken, is abelian. Correct me if it's non-abelian.
@KarlKronenfeld hmm does it help that each intersection point of the structure, amounts to solving a polynomial equation of the form $x^s=(1-x)^s$? That is to say, each point of the structure represents a specific algebraic object. Follow?
Well, if there is a different definition in regards to manually paired groups, we either weren't taught it or I simply don't remember it. There is a chance that a subtle difference was glossed over by either me or the professor.
@Alucard This reminds me, I was thinking of writing a novel set in a futuristic society where people lend their brains to their employer while they sleep; so work is replaced by working during sleep. This frees up time, but negatively affects isomniacs etc.
What are the differences between empty set, zero set and null set?
If i'm right empty set and null set is the same which is {}
but zero set is {0} ?
Is there a version of set theory that allows the existence of a set that does not admit the empty set as a member? I.e., reject the axiom $A\cup \emptyset = A$
