About the Rules & Moderation category (Part 1)

Announcements Rules & Moderation
#19327

that seems pretty straightforward just from the definition of F

#19328

Right but I don’t know what closed or open mean…

#19329

Or I didn’t. Until just now

1 Like
#19330

an arbitrary union of open sets is open

1 Like
#19331

that’s all you need to know

1 Like
#19332

this is presumably under the standard topology of R, where open sets are defined as exactly what you’d think they’d be

1 Like
#19333

for future reference, under a given topology of a set S, the empty set, S itself, and any arbitrary unions or finite intersections of open sets are open

2 Likes
#19334

I remember a bit from the topology final problems

#19335

just for the record because it never came up i was Lying

1 Like
#19336

i forget that a null set is technically something different from the empty set in measure theory

quizbowl is at times pedantic about this

2 Likes
#19337

anyway a closed set just being the complement of an open set means that closed and open are not mutually exclusive. say hello to clopen sets!

(in particular, an immediate consequence of the definition of topologies is that for any given topology on a set S, the empty set and S itself are both clopen sets)

1 Like
#19338

I kind of don’t understand how to explain the complement is open…

#19339

the complement of F is just that union

#19340

I know what hte completement is. Wait let me think

1 Like
#19341

which is straightforwardly a union of open sets

#19342

The “straightforwardly” part is what I’m missing

#19343

under the standard topology in R any set of the form (a, b) is an open set

1 Like
#19344

Oh that’s the interval. Right

1 Like
#19345

I forgot it was. An interval. Okay. Thank you

#19346

B should just be:

b answer

Suppose I is an interval in F that contains more than one element. Pick two distinct elements of F; we will call them a and b; WLOG assume a < b. Because the rationals are dense in the reals, there exists q in the rationals such that a < q < b. So, q is in F.

However, because r_n is a sequence that contains all the rationals, there is some k such that r_k = q. The interval (r_k - (1/2^k), r_k + (1/2^k)) = (q - (1/2^k), q + (1/2^k)) contains q. So, the union (write the union expression here) contains q, so F does not contain q.

This is a contradition, so our assumption is false, so all intervals in F contain exactly one element.

1 Like