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Stream: theory: mathematics

Topic: Basic topology

Julius Hamilton (Apr 13 2024 at 22:33):

I’m trying to understand some basic ideas in topology.

Topology is often presented as a way to “talk about space”, but it can be studied strictly algebraically.

You have a set of subsets, which are “closed” under the operations of union and intersection.

However, union and intersection can be defined in terms of each other. $A \cup B := \{x | x \in A \vee x \in B\}$. I think set comprehension can be expressed in terms of an implication arrow: $(x \in A \vee x \in B) \implies x \in A \cup B$. We can use De Morgan’s law to show that $\neg (x \in A) \wedge \neg (x \in B) \implies \neg (x \in A \cup B)$. Negate both side to get $\neg (\neg (x \in A) \vee \neg (x \in B)) \implies x \in A \cup B$. That says, “if it is not the case that $x$ is neither in $A$ or $B$, then $x$ must be in either $A$ or $B$”.

Now, I believe you can also get rid of the negation operator using a similar strategy. These are the arguments behind the idea that a Boolean algebra can actually be defined in terms of a single binary operator.

So, when a topology is viewed as an algebraic structure, can it be expressed in terms of a single law of composition on a set of elements?

Jean-Baptiste Vienney (Apr 14 2024 at 15:06):

I guess you're thinking to the definition using only the NAND? With a law of composition on a set of elements, you have more flexibility because of the large size of your input compared to a binary operator.

You could look at an operator which takes in input a family of subsets $(X_{i,j})_{I \in I, j \in J_i}$ of a set $X$ and do this:

$\textbf{m}(X_{i,j})=\underset{i \in I}{\bigcup}\Big(\underset{j \in I_j}{\bigcap}X_{i,j}\Big)$

But it's cheating because you have in fact an infinite family of operators depending on the sets $I$, $J_i$.

On the other hand, a more algebraic definition of a topological space is given by the Kuratowski axioms. It uses a unary operation $\mathcal{P}(X) \rightarrow \mathcal{P}(X)$ which corresponds to taking the closure of a subset, the binary union and the empty set, but also the symbols $\subseteq$ and $=$. Note that when you use $=$, you can replace it by two uses of $\subseteq$.

You can probably use an operator
$\textbf{d}(A,B,C,D)=\overline{(\overline{A} \cup \overline{B})} \cup \overline{(\overline{C} \cup \overline{D})}$
on $\mathcal{P}(X)^4$
where I use the usual notation $\overline{A}$ for the closure instead of the $\mathbf{c}(A)$ of Wikipedia.

Now I believe you can write the axioms for a topological space using this operator, the empty set and the symbol $\subseteq$.

It is not extremely satisfying because it is still using three different symbols, compared to two ones for a boolean algebra which are the nand and the equality. Moreover the important one takes four inputs compared to two for a boolean algebra.

Jean-Baptiste Vienney (Apr 14 2024 at 15:16):

Note that we probably can't use exactly "a law of composition on a set of elements". As you know, everything is in terms of subsets of a fixed set so the primitive objects are not exactly elements but these subsets of a fixed set. It is not exactly like in algebra.

Jean-Baptiste Vienney (Apr 14 2024 at 15:29):

Well, by looking more at the Wikipedia page, I now think we can maybe do better.

Jean-Baptiste Vienney (Apr 14 2024 at 15:29):

Screenshot-2024-04-14-at-11.28.44AM.png

Jean-Baptiste Vienney (Apr 14 2024 at 15:31):

You can maybe define a single operator of the type $\mathcal{P}(X)^n \rightarrow \mathcal{P}(X)$ for some nonnegative integer $n$ and rewrite this equation in terms of this operator only.

Jean-Baptiste Vienney (Apr 14 2024 at 15:35):

But I think we will still have to put $\emptyset$ in the entries of our operator in order to reobtain all the axioms of a topological space, as it is required when you use this single axiom if you look at this Wikipedia page.

Jean-Baptiste Vienney (Apr 14 2024 at 15:36):

Anyway, this is probably the best we can do to answer your question.

Jean-Baptiste Vienney (Apr 14 2024 at 15:58):

The final answer should be this:

If you define the operator $\mathbf{e}:\mathcal{P}(X)^4 \rightarrow \mathcal{P}(X)$ by:

$\mathbf{e}(A,B,C,D)=\big(A \cup \overline{A} \cup \overline{\overline{B}} \cup \overline{C \cup D}\big) \backslash \overline{\emptyset}$

then the preceding equation can be written:
$\mathbf{e}(A,B,\emptyset,\emptyset)=\mathbf{e}(\emptyset,\emptyset,A,B)$

So that a topological space can be defined as a set $X$ together with an operator $\mathbf{e}:\mathcal{P}(X)^4 \rightarrow \mathcal{P}(X)$ such that:
$\mathbf{e}(A,B,\emptyset,\emptyset)=\mathbf{e}(\emptyset,\emptyset,A,B)$

It looks funny but I think it's correct (tell me if I made a mistake!).

Jean-Baptiste Vienney (Apr 14 2024 at 16:06):

Hmm, it doesn't really work because you can't get back to the closure by any way but it must be fixable by allowing more inputs.

Jean-Baptiste Vienney (Apr 14 2024 at 16:13):

This one must be correct:

Define the operator $\mathbf{f}:\mathcal{P}(X)^5 \rightarrow \mathcal{P}(X)$ by:
$\mathbf{f}(A,B,C,D,E)=\big(A \cup \overline{B} \cup \overline{\overline{C}} \cup \overline{D \cup E}\big) \backslash \overline{\emptyset}$

Now Pervin's equation can be rewritten:
$\mathbf{f}(A,A,B,\emptyset,\emptyset)=\mathbf{f}(\emptyset,\emptyset,\emptyset,A,B)$

Jean-Baptiste Vienney (Apr 14 2024 at 16:14):

So that a topological space can be defined as a set $X$ together with an operator $\mathbf{f}:\mathcal{P}(X)^5 \rightarrow \mathcal{P}(X)$ such that:
$\mathbf{f}(A,A,B,\emptyset,\emptyset)=\mathbf{f}(\emptyset,\emptyset,\emptyset,A,B)$

for every $A,B \in \mathcal{P}(X)$.

Jean-Baptiste Vienney (Apr 14 2024 at 16:16):

To get back to a topological space in terms of closure, define:
$\overline{A}=\mathbf{f}(\emptyset,A,\emptyset,\emptyset,\emptyset)$
and then follow what's said on the Wikipedia page about Pervin's axiomatization and Kuratowski axioms in general to get back to the usual definition.

Jean-Baptiste Vienney (Apr 14 2024 at 16:18):

I'm still a but surprised about what I write but I invite you to verify everything and tell me if it doesn't work!

Jean-Baptiste Vienney (Apr 14 2024 at 16:20):

Damn, there is still a mistake.

Jean-Baptiste Vienney (Apr 14 2024 at 16:21):

Because $\mathbf{f}(\emptyset,A,\emptyset,\emptyset,\emptyset)=\overline{A} \backslash \overline{\emptyset}$

Jean-Baptiste Vienney (Apr 14 2024 at 16:25):

But maybe it's correct because you should get $\overline{\emptyset}=\emptyset$ at some point. I don't know.

Jean-Baptiste Vienney (Apr 14 2024 at 16:55):

Brr I think it's all very confused but it's maybe or maybe not possible to get a definition in this style which works. I'll think about this.

Jean-Baptiste Vienney (Apr 14 2024 at 18:02):

Now I think none of this makes sense but I tried :upside_down:

Jean-Baptiste Vienney (Apr 14 2024 at 18:07):

Maybe I'll have an "epiphany" later :sweat_smile:

Max New (Apr 14 2024 at 20:15):

Maybe not what you're looking for but (classically) you can define compact Hausdorff spaces as algebraic structure on sets that specifies what every ultrafilter converges to (https://ncatlab.org/nlab/show/compactum#algebras). You can weaken this definition to get all topological spaces by only giving a relation that says which ultrafilters converge to which points (https://ncatlab.org/nlab/show/relational+beta-module)

Julius Hamilton (Apr 15 2024 at 02:05):

Jean-Baptiste Vienney said:

Maybe I'll have an "epiphany" later :sweat_smile:

Thanks, yeah I appreciate you writing all this out, it’ll give me lots of work through and think about

Julius Hamilton (Apr 15 2024 at 02:06):

Max New said:

Maybe not what you're looking for but (classically) you can define compact Hausdorff spaces as algebraic structure on sets that specifies what every ultrafilter converges to (https://ncatlab.org/nlab/show/compactum#algebras). You can weaken this definition to get all topological spaces by only giving a relation that says which ultrafilters converge to which points (https://ncatlab.org/nlab/show/relational+beta-module)

Yeah that’s great. I read up a little bit on Stone duality which sounds relevant.