Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: learning: questions

Topic: Is there any function satisfying the following properties?

সায়ন্তন রায় (Feb 10 2024 at 07:29):

Let $\mathscr{L}$ be a non-empty set and let $\Gamma\cup \Delta\subseteq \mathscr{L}$. Does there exist any function ${\mu_\Gamma^{\Delta}}:\mathscr{L}\to\{0,1\}$ such that for all $\Sigma\subseteq \mathscr{L}$ ${\mu_\Gamma^{\Delta}}(\Sigma)\subseteq \{1\}$ iff $\Gamma\subseteq \Sigma\subseteq \Delta$? (This seems to be a very natural generalisation of the characteristic function, but I can't seem to find anything in literature , nor can I come up with something on my own.)

Todd Trimble (Feb 10 2024 at 13:54):

I'm looking at the question as literally written. The condition $\Gamma \cup \Delta \subseteq \mathcal{L}$ just says that $\Gamma, \Delta$ are subsets of $\mathcal{L}$.

Assuming $\mu_\Gamma^\Delta$ exists, since $\mu_\Gamma^\Delta(\emptyset) = \emptyset \subseteq \{1\}$, the biconditional would force $\Gamma \subseteq \emptyset \subseteq \Delta$. So $\mu_\Gamma^\Delta(\Sigma)$ exists only when $\Gamma = \emptyset$. For $\Gamma = \emptyset$, take $\mu_\Gamma^\Delta$ to be the characteristic function of $\Delta$. (and I believe this is the only example).

But is this really what you meant to ask?

সায়ন্তন রায় (Feb 10 2024 at 14:18):

Yes, you are right. I need to exclude the case of $\emptyset$. I assumed $\Sigma$ to be nonempty @Todd Trimble.

সায়ন্তন রায় (Feb 10 2024 at 14:24):

I also assumed that $\Gamma\subseteq \Sigma$.

Amar Hadzihasanovic (Feb 10 2024 at 15:40):

Are you saying that the correct condition is $\mu_\Gamma^\Delta(\Sigma) \subseteq \{ 1 \}$ if and only if $\Gamma \subseteq \Sigma \subseteq \Delta$ assuming that $\Gamma \subseteq \Sigma$?

In that case this is just the characteristic function of $\Delta$ all over again, since the fact that $\mu_\Gamma^\Delta(\Gamma) \subseteq \{1\}$ forces $\mu_\Gamma^\Delta$ to be equal to 1 on every element of $\Gamma$, anyway.

Amar Hadzihasanovic (Feb 10 2024 at 15:43):

If you want the condition to include at least one $\Sigma$ which is a proper subset of $\Gamma$, then it's impossible, because $\mu_\Gamma^\Delta(\Sigma) \not\subseteq \{1\}$ would imply that there exists $x \in \Sigma \subset \Gamma$ with $\mu_\Gamma^\Delta(x) = 0$, and then also $\mu_\Gamma^\Delta(\Gamma) \not\subseteq \{1\}$.

সায়ন্তন রায় (Feb 10 2024 at 17:02):

I think I have messed things up. I wanted to write $\Gamma\subseteq \Delta$, not what I have written. Extremely sorry for all the confusion.

সায়ন্তন রায় (Feb 11 2024 at 04:53):

Nevermind. I think I figured out that it is not possible.

Morgan Rogers (he/him) (Feb 11 2024 at 16:35):

I think I agree: clearly any element outside $\Sigma$ must take the value $0$, but even if we send everything in $\Sigma$ to $1$, the function has no way to tell if an element of $\Gamma$ is missing from a subset.

সায়ন্তন রায় (Feb 12 2024 at 10:45):

Yes. In fact, if we choose any $\Gamma$ which is not a singleton set then we will obtain a contradiction as follows. Suppose such a function exists. Choose any $\Sigma\subseteq \mathscr{L}$ such that $\Gamma\subseteq \Sigma\subseteq \Delta$. Since $\Gamma\subseteq \Sigma\subseteq \Delta$, $\mu^{\Delta}_{\Gamma}(\Sigma)\subseteq \{1\}$ - implying that $\mu^{\Delta}_{\Gamma}(\beta)=1$ for all $\beta\in \Sigma$. But $\mu^{\Delta}_{\Gamma}(\beta)=1$ implies that $\Gamma\subseteq \{\beta\}\subseteq \Delta$. This is a contradiction since $\Gamma$ is not a singleton set.

Amar Hadzihasanovic (Feb 12 2024 at 19:41):

@সায়ন্তন রায় a few lines ago I gave you a proof which has yours as a special case (I showed that your condition leads to a contradiction as soon as we apply it to a proper subset $\Sigma$ of $\Gamma$, which specialises to the case where $\Gamma$ has more than one element and $\Sigma$ is a singleton), so it is a bit rude to present “your” argument without acknowledging it at all.

সায়ন্তন রায় (Feb 13 2024 at 05:29):

@Amar Hadzihasanovic, I am really sorry if it came that way. I had no such intention. I thought that to anyone going through this thread, it should be obvious that this argument is a particular case of the one you gave (mine is just a more detailed version of the same). In any case, since you were going to be acknowledged anyway in the paper I am currently writing, I forgot to acknowledge you here. I will keep this in mind in future. Many thanks and sorry again.

Amar Hadzihasanovic (Feb 13 2024 at 07:15):

No worries at all, I am not asking for "credit" and it's completely up to you whether you want to acknowledge me in the paper, from the thread it just looked to me as if my message was not acknowledged at all. Anyway I'm sure it was just a miscommunication. Cheers :)