Let $ D $ be the open unit disc in $ \mathbb{C} $ and $ H(D) $ be the collection of all holomorphic functions on it. Let $$ S= [f\in H(D):f(\frac{1}{2n})=\frac{1}{2n}\forall n\in \mathbb{N}]$$ and $$ T=[f\in H(D):f(\frac{1}{2n})=f(\frac{1}{2n+1})=\frac{1}{2n} \forall n\in \mathbb{N}]$$.
Th...
"Several of the motivating examples assume an introductory graduate course in algebraic topology but may be skipped over by the reader willing to accept that such a motivation exists. An exception is the last section (section 10.9) which requires some familiarity with point-set topology."
It has its own open problems. The easiest to explain are existence problems, like can you always find this extra structure on a manifold, a certain type of the extra structure, etc.
it's been through several versions, one which connects to a graph db, but I scrapped that idea
Right now the design is going to try to incorporate diagram transformation rules
without worrying about "adhesive categories" and gluing
for instance, composition can obviously be applied everywhere in-place even in a complex diagram, where as you don't always get commutativity if you take a product and glue it in just anywhere, so I'll account for that with an option (or something)
It's go to understand and parse a simple variable format
It has auto-indexing now, so when you create a node with $X_1$ you can auto-index on either $X$ or $1$
No, diagrams are finite
clearly
but when a scene node is labeled "C" it's really just saying the diagram it contains is in that category
It's got to do a complicated graph-matching with the latex strings and parsing out variables, etc, but there's networkx lib for graph searching, and it can always just be done with brute force, as these diagrams are usually no more than 100 nodes
Prove that every convex subset of $\mathbb{R}^k$ is connected. (don't solve this for me, let me try first)
what I have done so far:
I am using the result: A convex function on $(a,b)$ i.e. a function such that $f(tx+(1-t)y) \leq tf(x)+(1-t)f(y)$ is continuous, where $t \in (0,1)$ and $a<x,y<b$ .
Now, supposing that $E$ can be written as the disjoint union of two separated subsets $A$ and $B$ of $E$ such that $A \cup B =E$. We pick a point $\mathbf{a} \in A$ and a point $\mathbf{b} \in B$. Now, let $\mathbf{a} = (a_1,a_2,...,a_k)$ and $\mathbf{b} =(b_1,b_2,...,b_k)$. Now we break the each individual components of the point in three categories : $P=\{(a_i, b_i) : a_i>b_i\}$ and $Q=\{(a_j, b_j): a_j<b_j \}$ and $S=\{(a_k,b_k) : a_k=b_k \}$ [we compare the $n$ th position of $\mathbf{a}$ to the $n$th position of $\mathbf{b}$].
in the end I realized that I am getting into a wrong proof
There's a general approach to prove certain things are connected which is basically the path-connectedness idea but doesn't use paths explicitly. Let me know when want to hear it
so, if I can somehow manage to prove that every path connected set is connected, then a convex set will also be connected. Since for any two points $x, y $ , there is always a straight line (a continuous curve) that joins the two (via a specific kind of path)
or, more formally, $f:[0,1] \to A$ such that $f(t)=ty+(1-t)x$. where $f(0)=x$ and $f(1)=y$
Using that same kind of function f from the open interval [0,1] to any path connected set X you can also prove that any path connected set is connected. It's actually a very nice proof.
so, when it is the codomain set of a continuous function whose domain is a connected set, then to maintain the "connectedness", it takes either one of the values
Suppose $E \subset X$ is a disconnected set but it is arc connected. So, we can find a pair of non empty open sets $A$ and $B$ such that $A \cup B= E$.
Pick $x$ and $y$ respectively in $A$ and $B$. But, being path connected, we can find a continuous function $f:[0,1] \to E$ such that $f(0)=x, f(1)=y$.
Now, for an arbitrary continuous function $g$:
$I$ connected $\implies g(I)$ connected.
So, contrapositively $g(I)$ not connected $\implies I$ not connected.
Here, $E$, i.e $f([0,1])$ is not connected, but $[0,1]$ is connected. A contradiction.
@AkivaWeinberger I have not studied much about the universal set, but removing foundation is absolutely necessary. You might need to use urelements (thus depending on how they are implemented, may blow up the axiom of extensionality as now you have a lot of things that cannot contain elements and yet are different from the empty set)
New Foundations for example, use a hierarchy of types and urelements to produce the universal set
Hmm...
$\{x\in V\mid x\notin x\}$ sounds like the set of all well founded sets and it obviously does not contain $V$ because $V$ is an element of itself (along with many non well founded sets) hence does not satisfy the predicate on the right
So this set will actually be smaller than $V$ I think (assuming foundations is nuked)
$\{x\in V\mid x\notin x\}$ should be an element of $V$ unless my brain is not working
I don't think in any way you can end up with something like:
$\{x\in V\mid x\notin x \land x\in x\}$ which is the Russel set
That is, I suspect $\{x\in V\mid x\notin x\} \cap \{x\in V\mid x\in x\} = \varnothing$
o wait, then we have: $\{x\in V\mid x\notin x\} \cap V = \varnothing$, I am not sure how to interpret this
yeah, I think blowing up foundation is not enough, the consistency of the Russell set is something that cuts much deeper than set axioms
New Foundations get around that problem because by making types a hierarchy, the Russell set cannot be a set
Meanwhile Wikipedia:
> ZFC does not assume that, for every property, there is a set of all things satisfying that property. Rather, it asserts that given any set X, any subset of X definable using first-order logic exists. The object R discussed above cannot be constructed in this fashion, and is therefore not a ZFC set. In some extensions of ZFC, objects like R are called proper classes.
So blowing up foundations will not automatically allow you to have the Russell set in $V$
I am a beginner in mathematics and I was reading a text on Set Theory that talked about how Zermelo's Axiom of Selection "solves" Russel's Paradox.
I understand that the the axiom does not allow constructions of the form
$$\{x \:: \text S(x) \}$$ and only allows$$\{x \in \text A \:: \text S(x) \...
The Russell set obeys the inclosure schema, which is purely of logical consequences. Set theories are meta systems and thus they alone are not strong enough to block the Russell paradox (Not sure if I am using the terms correctly here as my logic is still not very good)
One way to restrict axiom of specification is to ban the use of negations in the predicates. This immediately forbids $x \notin x$ and avoid the Russell paradox. The resulting set theory is called positive set theory.
If you want a Russell set to be explicitly included in the universal set, I am afraid you might have the change the underlying logic system to avoid the inclosure schema from causing trouble
yes. It was a mistake. $E$ is not the range, rather it is a codomain set.
As to explain why $f[0,1]$ is not connected:
[first of all, I forgot to mention that $A$ and $B$ are such that $A \cap B = \emptyset$ ]. We consider the sets $P =f[0,1] \cap A$ and $Q =f[0,1] \cap B$. Clearly, $P \subset A$ and $Q \subset B$ and $\overline{P} \subset \overline{A}$ and $\overline{Q} \subset \overline{B}$. Therefore, $\overline{P} \cap Q = \emptyset$ and $\overline{Q} \subset P = \emptyset $ [since $\overline{A}\cap B= \emptyset$ and $\overline{B} \cap A =\emptyset$] . Further, $P \cup Q = f[0,1]$…
@SubhasisBiswas Sure, OK. All of this is too complicated though. You can directly argue $f^{-1}(A)$ and $f^{-1}(B)$ are disjoint open subsets of $[0, 1]$ such that their union is all of $[0, 1]$.
They are both nonempty as well, because $0 \in f^{-1}(A)$ and $1 \in f^{-1}(B)$. But $[0, 1]$ is connected, so this is an impossible scenario
Continuity means preimage of open set is open, that's all
ISI is a shit place and "X is/isn't worthy of ISI" is a garbage judgement thereof which only elitist fools who don't actually accomplish much academically keep saying in quora or whatever
Brightness is immaterial just keep learning/enjoying learning
@SubhasisBiswas Sounds like a horse-before-carts question. Loops are paths $\gamma : [0, 1] \to X$ with $\gamma(0) = \gamma(1)$, i.e., "same initial and terminal point"
@BalarkaSen I am. It is astonishing to me how much I am learning every day. This "path connectedness" is a way to view the world. Intuitively, as I have interpreted it, it acts sort of a bridge between any two points in a set.
perhaps there are some holes in my view, but for now, this is what I think of it
Homotopy is a different ballgame. You say two paths $\gamma_1, \gamma_2 : [0, 1] \to X$ with $\gamma_i(0) = p$ and $\gamma_i(1) = q$ for $i = 1, 2$ are homotopic if you can "continuously slide $\gamma_1$ to $\gamma_2$" while keeping the endpoints $p, q$ fixed.
@BalarkaSen ok. So, as I keep sliding the curve smoothly to another (i.e. reshaping the curve into another, but in a smooth way), the initial "area" (in two dimensional world) must be covered smoothly...
@SubhasisBiswas Yes, but you can use connected sets as "bridges" as well. If $X$ is a topological space such that for every two points $x, y \in X$ there is a connected subset $C_{xy}$ of $X$ containing both $x$ and $y$ then $X$ is connected. You can prove this.
By the definition of path connectedness, for any two $x, y \in \mathbb{R}$, there is a continuous function $f:[0,1] \to \mathbb{R}$ such that $ f(0)=x$ and $f(1)=y$. [ for example, consider the function $f(t)=(1-t)x+ty$, $t \in [0,1] \ \ $ ]. Now, map $\mathbb{R} \setminus [0,1]$ to $0$.
Okay, consider the graph as a set $G \subseteq \mathbb{R}$. The graph is path connected. So, we can indeed find a continuous function $f_{\{x,y\}}:[0,1] \to \mathbb G$ such that for any two $x, y \in G$, $f(0)=x$ and $f(1)=y$. Now, for a bounded set, let $\sup G =M$ and $\inf G= m$. We consider $f_{\{m, M\} }$
We map every point in $\mathbb{R}\setminus [0,1]$ to some $p \in G$
I don't need all this "can collections contain themselves" and "what about the collection of all collections" riddle bullsh*t
("And if they can contain themselves, what about the collection of all the ones that don't")
…The word collection has stopped sounding like a word
Maybe jamais vu is what allowed set theory to develop. The more you say "set", the less it sounds like a real word, the better you feel about never actually defining the things
Let $(X,\tau)$ be a topological space. Then we know some conditions under which $(X,\tau)$ is metrizable (see for example this and this). It is also clear from these theorems that not every topological space is metrizable.
However, I am wondering whether the same is true for pseudometric spaces ...
Given two vector fields $x,y$ and their flows $\varphi_x^t,\varphi_y^t$, is there a simple expression for the flow of the commutator, ie for $\varphi_{[x,y]}^t$?
the simplest I've got is that its the flow associated to the vector field $[x,y]_p=\frac12\frac{d^2}{dt^2}( \varphi_{y}^{-t}\varphi_x^{-t}\varphi_y^t\varphi_x^t )(p)\lvert_{t=0}$
If your vector fields are left-invariant vector fields on a Lie group then there's the Baker-Campbell-Hausdorff formula which I think gives you an explicit answer
I encountered this some time ago when thinking about the Lie bracket
Suppose, $X$ is not connected. Then there exists a pair of non-empty open subsets $A$ and $B$ of $X$ such that $A \cup B =X$ and $A \cap B = \phi$. Choose $x \in A$ and $b \in B$. By hypothesis there is a connected subset $C_{xy}$ of $X$ that contains both $x$ and $y$.
Consider $P= C_{xy} \cap A$ and $Q =C_{xy} \cap B$ . Now, $P \subset A$ and $Q \subset B$ $\implies \overline{P} \subset \overline{A}$ and $\overline{Q} \subset \overline{B}$ .
Now, $\overline{A} \cap B = \emptyset$ and $\overline{B} \cap A = \emptyset$. Therefore, $\overline{P} \cap Q= \emptyset$ and $\overline{Q} \cap P …
@SubhasisBiswas I don't know what you're doing with the closures flying around everywhere. $P$ and $Q$ are open in $C_{xy}$ by definition of subspace topology, nonempty because $x \in P$, $y \in Q$, and $P \cup Q = C_{xy}$ as you rightly observed. That's all!
Note that path-connected => connected falls out as a corollary from this, by the line of reasoning you were progressing along previously
For any $x, y \in X$, let $\gamma : [0, 1] \to X$ with $\gamma(0) = x, \gamma(1) = y$. Let $C_{xy} = \gamma([0, 1])$. That's a connected subset containing $x$ and $y$.
Well you already had the right circle of ideas so I just disclosed it instead. To compensate, here's another exercise: Prove that finite (actually, do it without this adjective as well) product of connected topological spaces is connected.
Thus unlike in non relativistic quantum where the "basis states" have clear physical meaning under some observables, some of the fields $\phi$ corresponds to fields in new physics, and hence we don't know what they are until we discovered them
infinity is when we can guarentee we can always discover new physics regardless of how far humans are into the future civillisation
For else by proof by contradiction, there exists some maximum finite number $M$ where we will stop finding new physics
Well, $\aleph_0$ is a must, for it is where the whole notion of infinity begins. Being the hardest thing to get to from below (you cannot reach it by applying hartog number construction a finite number of times on any finite number, and with the exception of NF (which is basically a cheat since it is derived from the universal set) no foundations I knew of can construct it from below), being able to touch $\aleph_0$ will be a milestone
I have no idea, it really depends on whether there exists physical infinity, which as far I am aware, it is still an open question
but suppose it does not exists, then the natural question to ask is what is a finitistic proposition that form a one to one correspondance with the definition of e.g. $\aleph_0$
This is because if physical infinity does not exist, then either it is completely idealistic, or that it is a shorthand for something we do not quite understand, and finding what it is actually a shorthand for can be very illuminating in understanding the concept of at least what it means to be countable
So,if physical infinity does not exists, then obviously you cannot touch it in the literal sense, but you may be able to touch it figuratively by unravelling the underlying notion on what it really means
