Examples of topologies and closed sets -- Topology video 2

[@alexandreocadiz9967]

Thx for the video. Looking forward to the episode where we will transform a teacup into a donut.

[@AKnipp412]

I’ve been waiting for a series on Topology. Thank you!

[@radadadadee]

In the example at 18:00, one could also set T to be the collection of closed intevals [-n, n] and it would still constitute a topology. Then each closed interval [-n, n] would be an open set. So is it open or closed?

[@radadadadee]

I guess [-n, n] is a closed set under the standard topology on R, but it's an open set under this non-standard "concentric" topology on R?

[@methatis3013]

You are correct

[@gp-ht7ug]

Two videos today!! 👏👏

[@Alan-zf2tt]

Loving this topic! But a confusion @ 15:06 my perception is flipping between ordered pairs eg (-2, 2) and open intervals ]-2, 2[

[@plislegalineu3005]

the ]a,b[ notation is, I think, French, and generally the world uses round brackets

[@Alan-zf2tt]

@plislegalineu3005 this may be true but my perception seems to take fright at two 2-tuples indicating a range of values 🙂

[@methatis3013]

​@plislegalineu3005 idk about the rest of the world, but I think pretty common notation is also or 〈-2, 2〉(similar to scalar product)

[@Alan-zf2tt]

@@methatis3013 I think I may have to remain in a minority of one. It seems to me the reversed square bracket does an excellent job with limited ambiguity tho

[@methatis3013]

@@Alan-zf2tt downside: it looks hella ugly lol

[@GugasSoares]

professor could you perhaps show some visualizations of topologies? just for providing a more concrete intuition

[@Alan-zf2tt]

A mmm @ 3:00 or thereabouts. Case 1: U intersect V is empty set therefore is contained in T (?) Case 2 as demonstrated. Plus - a maybe for Case 2 based on a WLOG on intersection operator on arbitrarily indexed sets - change closing bracket on U(k) to U(k-1) then en bracket U(k) intersect U(k+1) Something like: U(1) chained intersection to U(k) intersection U(k+1) is identical to U(1) chained intersection to U(k-1) intersect U(k) chained intersection U(k+1). I am not too bothered about unique Ui at this stage and for the purposes of above proof I claim I can invent any number of identical Un to make the proof work. Basis: all the operations at this stage are proof based principles on generalities A WLOG as the indexing is general and may or may not be ordered on any criteria and may or may not be repetitive according to desire. Remark: this is a question I would ask in lecture besides U2 identical to U3 does not perturb initial assumptions nor results. And I can jiggle the Ui any way WLOG and almost abuse the poor Indexing set I

[@Alan-zf2tt]

@@radadadadee Happen so, my friend. Happen so. Consider 4528 in decimal. Replace powers of ten in (assumed) decimal system with powers of ascending primes ordered so as to increase from right to left. Unlike decimal system it is a closed numbering system in sense that any prime to power zero adds a "1" and that can be awkward. Damn, bloody awkward really. So maybe we might need a symbol such as ¥ to denote "No power here" Anyway, in that awkward, closed numbering system, call it FrEd, 4528 in decimal returns 5792 in FrEd numbering system. FrEd seems to have advantage of turning p/q rationals into plain old p rationals as negative powers of primes can be helpful. Example rational 1/2 in decimal 0.5 is (-1) in FrEd and 1/3 becomes 0.3r becomes (-1)0 and see! It is awkward, bloody damned awkward 🙂 Further suggesting a limitation that 1 in decimal is for ever 0 in FrEd or ¥0 to disambiguate as 0¥¥¥0 would be vbery very confounding

[@Alan-zf2tt]

I am thinking about extending FrEd to FrEd1 by including 1 as a prime. It removes ambiguities of prime to power zero representing far to many units of 1 most of which are redundant. It also means itegrity of quotient as p/q can be maintained rather than shrink p/q to p using negative indices on powers of primes/ Keeping them in denominator seems equally helpful as does permitting negative indices in numerator. Who says math has to be super serious and super intense and no fun?

[@Alan-zf2tt]

hmmm or maybe not as presentation of powers of primes requires primes to power zero to be 1 anyway - slap on forehead - silly me! So in this case we can say one is identically equal to zero?

[@psychSage]

You say in theorem, that singletons are included in some topology, so they are open, and then that they are closed. Where's a mistake?

[@jimallysonnevado3973]

singletons can be open and closed at the same time in some topologies there‘s no mistake

[@阿波羅-y9f]

Singletons can be open in some topological space, but they are only closed but not open in the standard topology on R.

[@martinhazard2025]

A topological space is a set X? no. A topological space is an ordered pair.

[@methatis3013]

"set X, together with a collection of subsets T..." This essentially means an ordered pair