On Definition 2.3 (F_del and G_sig):
uncountable intersection example:
irrat = bigcap { { r }^c : r a rational }
are a countable intersection of open sets
Open question (for us):
is there any set that is neither G_del nor F_sig?
2.2 Neighborhoods and closures
A^circ means "interior of A" means largest open set within A.
--------X-----------|
_______ |
| . | V
--------
- every set has a boundary but the boundary might be the empty set
impose standard topology over X = [-4, 10]
WRONG: boundary of X = { -4, 10 }
and X is clopen
Errattum: Your space X always has empty boundary as otherwise you'd
have a point that non-trivially intersects with stuff outside the space,
but by definition such stuff does not exist.
We KNOW all open sets DO NOT CONTAIN their boundary points.
Refer also to: https://www.amazon.com/Counterexamples-Topology-Dover-Books-Mathematics/dp/048668735X/ref=asc_df_048668735X/?tag=hyprod-20&linkCode=df0&hvadid=312165853622&hvpos=&hvnetw=g&hvrand=8743399671892861203&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9030244&hvtargid=pla-449277484053&psc=1&tag=&ref=&adgrpid=60258872297&hvpone=&hvptwo=&hvadid=312165853622&hvpos=&hvnetw=g&hvrand=8743399671892861203&hvqmt=&hvdev=c&hvdvcmdl=&hvlocint=&hvlocphy=9030244&hvtargid=pla-449277484053
aka the counterexamples in topology book.
REGARDING perfect subspaces:
Consider the real line.
{ 4 } U [0, 1] is not perfect as shown by considering (3.5, 4.5)
if 4 were a limit point in A
there would exist a sequence, in A, converging to 4, containing stuff other than 4
implying that there are elements of A arbitrarily close to 4
implying that every neighborhood of 4 in A contains a point of A other than 4
namely one of those arbitrarily close things from the sequence
{ 4 - 1/n | n in Nat }
will contain 4 - 1/2 = 3.5
but there is no sequence WITHIN that sequence which conveges to 3.5
hence 3.5 is not an accumulation point
hence, the sequence is not perfect
( ( A B ) C )
in this one, ( A B ) is perfect
because every neighborhood of A has B and vice versa
so in conclusion they don't need to be "big"
BTW If you ever have someone ask for a good intro to pre-orders and such things,.
chapter 1 of this text is very very nice (according to me, Max): https://www.logicmatters.net/resources/pdfs/Galois.pdf
a in A
b in B
(a, b) in A x B
Munkres writes a x b in A x B
a x b x c is in A x B x C
as opposed to ((a, b), c) or (a, (b, c))
(a, b) in A x B and c in C
(a, b) x c = (a, b, c)
looks kind of weird
compared to a x b x c
also kind of nice in power-set spaces which you do a lot of in topo.
so basically the notation handles the obvious isomorphism for you.
{ (ra, rb) | ra, rb are rational }
(-pi, e)
Union (-1 * first N digits of pi | N in nat, similar for e)
Hence, we proved (ish) that R1 is 2nd countable.
C(X) is the space of continuous functions X -> X
A function f : X -> Y on topological spaces (X, T_X) and (Y, T_Y)
is considered continuous if and only if, for all O_Y in T_Y,
f^-1 O_Y is in T_X.
------
2.2.7
Consider the topology (R, T) where T is defined below.
T = {
A | x in A -> exists U open in R,
exists C countable,
such that U - C is a subset of A
}
A base for T is the set
B = { U - C where U is open in R and C is countable }
Only points in this space that converge are ones that are eventually constant.
0----- Continued on 17 July 22
Suppose { x_alpha } alpha_in_A is a net.
Let phi : Delta -> A be a function such that condition 2 holds.
Then { x_phi(d) } d_in_Delta is called a subnet of the original net on line 3.
phi
Delta ----------> A
n
A --------------> X
(1) Delta ----> A -----> X
<------
d |----------------> x
a such that n(x) = x
a < a' -> f(a) < f(a')
a in A then exists x in codomain(f) such that b > a ->
{ ()}
---- Cafe August 7 2022 ----
2.28 Theorem - Let f : X->Y be a function and x in X.
Let (X, T) and (Y, G) be the corresponding spaces.
TFAE:
I. f is continuous at x.
II. if a net x_a -> x in X, then f(x_a) -> f(x) in Y.
III. if a filter F -> x in X, then f(F) -> f(x) in Y.
PROOF.
Suppose f is continuous at x.
<-> For every open neighborhood V of f(x),
f^-1(V) is a neighborhood in X containing x.
Let F be a filter in X and suppose x is a limit point of F.
Therefore the closure of any s in F contains x.
Thus f(closure(s)) contains f(x) for each s in F.
Recall that f(closure(s)) <= closure(f(s)) as f is continuous.
Thus closure(f(s)) contains f(x) for each s in F.
But this suffices to show that f(F) -> f(x) in Y.
So we have proven that I -> III.
Suppose {x_a} is a net and x is a limit point of it.
Consider the filter F generated by this net.
It has the same limit points.
So x is a limit point of F.
So by prior logic, f(F) -> f(x) in Y.
We still need to show f(x_a) -> f(x) in Y.
NTS: Let V be an arbitrary nbrd of f(x), and a be arbitrary index.
We claim: exists b >= a s.t. x_b in V.
TODO - work on this more next time.
21st is next meeting
September 4 2022
---
Proof of 2.32 Lemma in Hitchhiker's Guide:
K is a compact subset of a T2 space
Consider J = { (O, V) : k is in K and O is an
open neighborhood of k and V is an open neighborhood of x
and O intersect V = emptyset }
Notice all these pairs must exist because the space is T2
Clearly the set of all the Os is an open covering of K
By compactness of K, there exists a finite subset of this set,
which is likewise an open covering of K
Take the finite subset of corresponding Vs, and take its
(finite) intsersection; this must also be open as finite
intersection of open sets is open, and must include x because each
V contains x
Call this new set V_final
Call the union of the finite collection of Os we came up with, O_final
Clearly O_final intersect V_final = emptyset
and O_final contains K and V_final contains x
QED
Claim: Compact subsets of T2 spaces are closed.
recall that arbitrary intersection of closed sets is closed.
Let K be a compact subset of a T2 space.
For each point x not in K, let (O_x, V_x) be a tuple such that O_x, V_x are both open sets, O_x contains K, and V_x contains x, and the intersection of O_x, V_x is the empty set.
Now take the intersection of all the V_x^c's for all the xs not in K.
This is an arbitrary intersection of closed sets that must contain all of K.
Indeed, this set cannot contain any x NOT in K, as to do so, it would need to contain an open neighborhood V_x, which cannot be the case
as it's part of the intersection with V_x^c.
So, this set doesn't contain anything not in K, and does contain K, therefore by excluded middle, the set is K.
QED.
Ansatz: A homomorphism of groups can be thought of as a set isomorphism.
Let (G, +) and (H, *) be groups.
Let f : G -> H be a homomorphism of groups.
So, importantly, f(x + y) = f(x) * f(y).
Consider the set G_zipped = { (x, y, x + y) : x, y in G }
H_zipped = { () }
Let h : G_zipped -> { (f(x), f(y), f(x + y)) | x, y in G } be the function
h(x, y, x + y) = (f(x), f(y), f(x + y))
= (f(x), f(y), f(x) * f(y))
= H_zipped
Because f is a homo. of groups, we were able to make
a function h which was obviously equivalent, where h
was a set isomorphism.
f(a) = x if and only if (x, 0, x) = h(a, 0, a).
G = (N, +)
H = ({ 0 }, +)
f(x) = 0
Conclusion: Max's Ansatz is wrong.
------------------------
2.10 Semicontinuous Functions
Let f : X -> R be a function.
f is "lower semi-continuous" if for each r in R the pre-image of
(-inf, r] is closed; and
"upper semi-continuous" if for each r in R the pre-image of
[r, inf) is closed.
X = [1, 3] in R
f (x) = x + 2
f(X) = [3, 5] which is not all of R
minimizer of f is f^-1(3) = 1
[-inf, inf] -> [-inf, inf]
minimizer = { -inf }
{0, 1, 2}
{ x >= 3}
f(x) = 0 if x <= 2 else 1
( ( { . . . . . } )
{ (.) ) . (.) . . . . . }
R = (-inf, inf)
[0, 1] = [-inf, inf]
Suppose for a contradiction x is a minimizer
Recall f : X -> R is our funciton, x in X
By supposition, f(x) is the smallest