infinite fold tensor product of universal enveloping algebra

Let $ \mathfrak a$ be a Lie algebra graded by the abelian semigroup $ S$ , then the universal enveloping algebra $ U(\mathfrak a)$ of $ \mathfrak a$ is $ S \sqcup \{0\}$ graded. I have the following questions.

  1. What is the definition of infinite fold tensor product ($ U(\mathfrak a)^{\otimes \infty}$ ) of $ U(\mathfrak a)$ and is this also $ S \sqcup \{0\}$ graded?
  2. If so, how to express the grade spaces of this infinite tensor product in terms of grade spaces of $ U(\mathfrak a)$ ?
  3. Is it a good notation $ U(\mathfrak a)^{\otimes \infty}$ ?

Thank you.

Which of the known alternative set theories is nearest in structure to this theory with a universal set and the complement of Russell set?

Before I’ll present the exposition of this theory, I’ll speak a little bit about the Mereological concept it is meant to catpure.

The idea is to work in Atomic General Extensional Mereology “AGEM”, one can think of it easily as a theory about collections of atoms, where atoms are indivisible objects, i.e. objects that do not have proper parts. The relation is an atomic part of is defined as:

$ \sf Definition:$ $ x P^a y \iff atom(x) \land x P y$ .

where $ P$ stands for “is a part of”, and atom(x) is defined as:

$ atom(x) \iff \not \exists y (y P x \land y \neq x)$

This atomic part-hood relation can be regarded, conceptually speaking, as an instance of set membership relation.

Now the following theory is a try to define a set theory by a strategy of mimicking properties of this atomic part-hood relation with the background theory being AGEM.

Notation: let $ \phi^{P^a}$ denote a formula that only use the binary relation $ P^a$ or otherwise the equality relation, as predicate symbols. The notation $ \phi^{\in|P^a}$ denotes the formula obtained by merely replacing each occurrence of the symbol $ “P^a”$ in $ \phi^{P^a}$ by the symbol $ “\in”$ .

Comprehension axiom schema: if $ \phi^{\in|P^a}(y)$ doesn’t have the symbol $ x$ occurring free, then all closures of:

$ $ \forall A [\exists x \forall y (y \ P^a \ x \leftrightarrow \phi^{P^a}(y)) \to \exists x \forall y (y \in x \leftrightarrow \phi^{\in|P^a}(y))]$ $

are axioms.

In order to complete this theory we add axioms of Extensionality, Empty set and Singletons:

Extensionality: $ \forall xy [\forall z (z \in x \leftrightarrow z \in y) \to x=y]$ .

Empty set: $ \exists x \forall y (y \not \in x)$

Singletons:: $ \forall A \exists x \forall y (y \in x \leftrightarrow y=A)$

This theory has a universal set, also has a set of all sets that are in themselves, however it doesn’t have complements; axioms of Set union and Power are there. There are separation axioms for formulas $ \phi^{\in|P^a}$ where $ \phi^{P^a}$ holds of at least one object. Similarily replacement axioms are granted if $ \phi^{P^a}$ formula replace atoms with atoms and of course is non empty.

The trick is that all formulas of the form $ x \not \in x$ , $ \exists x_1,..,x_n: \neg (x_1 \in x_2 \land…\land x_n \in x_1)$ ; $ x \text { is well founded }$ , $ x \text{ is a von Neumann ordinal }$ , etc.. all of those won’t have their $ \phi^{P^a}$ corresponding formulas hold of mereological atoms and so cannot be used in comprehension because there do not exist an object that has no atomic parts, since we are already working in AGEM.

Question: if one attempts to prove the consistency of this theory, which of the known alternative set theories have in some sense the nearest structure to this theory, other than positive set theory?

Can the “$” symbol be safely considered as universal when making a graphic depicting money?

Want to get a sense of any best practices of using the “$ ” dollar sign in the context of an image in order to depict a universal idea of money.

UPDATE: My task was to create an icon/graphic used in an e-commerce checkout process which depicted the idea of an ‘invoice’. It had to make sense in a global context (not just USD). Here is the exploration I had done: enter image description here

And here is the icon I ended up picking (without a symbol): enter image description here

Thanks for all the insights and weigh-ins everyone offered. VERY helpful!

Make Universal Clipboard fast and reliable

For me, the Universal Clipboard feature of Continuity has always been a bit spotty, but with newer hardware, it has become somewhat more reliable.

The problem is getting it to work consistently. I’ve read the Apple Support document, Use Universal Clipboard to copy and paste between your Apple devices which unfortunately is hopelessly simplistic.

Yes, I have Bluetooth, Wi-Fi, and Handoff enabled, and all of my devices are correctly signed into the same iCloud account. Other Continuity features (Handoff, notably) appear to be working just fine.

However, what I consistently experience now, between my iPad Pro 11” and my iPhone XS, is this: When I try to copy on one device and paste it on another, what happens is that it won’t work at all the first time, and then if I repeat the paste, the device on which I’m pasting will display an alert “Pasting from … ”, I’ll have to wait about 5 seconds, and then the text will be correctly pasted.

Previously I had seen this message appear when pasting a rather large text (say a page), but now it occurs repeatedly even when I’m only pasting only a few words. This is, however, not the case between the iOS devices and my 2018 MacBook Pro running macOS Mojave 10.14.5.

I’ve posted about this earlier — with different (older) hardware:

  • Is there any way to make Universal Clipboard in High Sierra reliable?

In the past, the most reliable fix for the Universal Clipboard not working at all — as reported on various sites — was to sign all devices out of iCloud account, restart, and log in to iCloud again on all devices. Yet as noted above, Universal Clipboard is working to some extent for me, while signing out of iCloud and signing in again is not as easy as it seems — many settings have to be re-established.

Are these sorts of problems par for the course with the Universal Clipboard? or is there a clear way to get it working quickly and reliably? All of my devices are running the latest versions of system software (macOS Mojave 10.14.5 and iOS 12.13.1).

Homology of the universal cover

$ k$ is a field. Let $ X$ be a connected pointed CW-complex such that the homology $ H_{n}(X;k)$ is a finite dimensional $ k$ -vector space for any $ n\in \mathbb{N}$ . Suppose that we have continuous mapping $ $ r: X\rightarrow K(\pi_{1}(X),1) $ $ from the space $ X$ to the Eilenberg-MacLane space $ K(\pi_{1}(X),1)$ inducing an isomorphism on fundamental group $ \pi_{1}$ . Suppose also that there is a mapping $ $ i: K(\pi_{1}(X),1) \rightarrow X $ $ such that:

  1. $ r\circ i= id$
  2. the homotopy cofiber of $ i$ is homotopy equivalent to a finite $ CW$ -complex.
  3. and $ H_{n}(r;k): H_{n}(X;k)\rightarrow H_{n}(K(\pi_{1}(X),1);k)$ is an isomorphism for any $ n\in \mathbb{N}$ .


My question is the following: Is the homology of $ \tilde{X}$ (the universal covering of $ X$ ) finite dimensional ? i.e. $ H_{n}(\tilde{X};k)$ is a finite dimensional $ k$ -vector space for any $ n\in \mathbb{N}$ ?


Given a family of hash functions in table form, how can I know whether it’s universal?

I’ve been given the following two families of hash functions:




enter image description here

Each family has three functions $ \{0,1,2,3,4\} \to \{0,1,2\}$ that can be seen in the tables above. For each family I need to decide whether it’s universal. I know that a family of hash functions $ F$ is called universal if for every $ x \neq y$ , $ \text{Pr}(f(x) = f(y)) \le \frac{1}{m}$ ($ f$ is a function in $ F$ ). However, I don’t understand how to calculate this probability. Should I calculate it for any one of the functions or for the whole family?

How to execute a dmg file using a custom universal keybaord shortcut?

It may sound strange but I need to run a specific dmg installer whenever I trigger a specific custom keyboard shortcut. Here on Sierra 12.6 I’ve assigned a custom keyboard shortcut to sleep/lockscreen mac using shellScript/appleScript.

I have no idea how the shell script would look like.

I’d be grateful for any help and participation.

How does Rogozhin’s (2, 18) universal turing machine work?

I am trying to understand Rogozhin’s (2, 18) universal turing machine by stepping through a simple 2-tag encoding that I believe should loop forever:

a -> aa 

For example, using an initial input of aaa:

aaa   aaa     aaa       .... etc 

Apologies for the extremely specific question, but it’s what I’ve narrowed my issue down to and I’ve been stuck for a while!

Following the instructions in part 10 / pp22 and page 6 I believe this system should be encoded as:

c⃖₁ c⃖₁ b  b  1  b  1  b >b  1  c  1  c  1  c |P2   |P1           |P0   |Ar   |As   |At   

Running this, however, results in termination rather than an infinite loop. Following the trace I have managed to identify something I can’t explain and that seems wrong, but have not been able to figure out a resolution.

Following the first stage of modelling:

On the first stage, the UTM searches the code P, corresponding to the code A, and then the UTM deletes the codes A, and A, (i.e. it deletes the mark between them)

if the head of the UTM moves to the right and meets the mark c, then the first stage of modelling is over. The UTM deletes this mark and the second stage of modelling begins

Gives the following trace:

c⃖₁ c⃖₁ b  b  1  b  1  b >b  1  c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b  b⃖ >1  c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b >b⃖  c₂ c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1 >b  b  c₂ c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b⃖ >b  c₂ c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b⃖  b⃖ >c₂ c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b⃖  b⃖  1⃖ >c  1  c  1  c c⃖₁ c⃖₁ b  b  1  b  1  b⃖  b⃖ >1⃖  1⃗  1  c  1  c 

At this stage, Rogozhin claims the tape should be:


Notice in particular R’At

R’ consists of 1⃖ and 1⃗ and the head of the UTM is located on the R’ in the state Q2

But to me, it appears that only Ar has been deleted!?

c⃖₁ c⃖₁ b  b  1  b  1  b⃖  b⃖ >1⃖  1⃗  1  c  1  c |P2   |P1           |P'0  |R'   |As!! |At   

I would expect something like:

c⃖₁ c⃖₁ b  b  1  b  1  b⃖  b⃖ >1⃖  1⃗  1⃖  1⃗  1  c |P2   |P1           |P'0  |R'         |At   

I have identified the following potential errors I have made, but have double checked each and have not been able to identify any:

  • Understanding of 2-tag system.
  • Encoding of 2-tag system.
  • Execution of rules (programming bug).
  • Formatting of trace.
  • Interpretation of trace.

Can anyone spot what I’m missing? Am hoping it’s something obvious!

Supplementary materials

  • Spreadsheet to generate tape
  • Ruby program I’m using to generate traces
  • Full trace showing that aaa terminates when I think it shouldn’t
  • Original motivation and how I ended up here in the first place