FNATIC'S FATHER

of eco frags

5-5 on a defensive map

so ez for loud kkkkkkkk

there was no eco frags to farm since fnatic lost pistols

4-3 on a defense sided map
less owning his ass rn

they are getting exposed by a good team from a good region

BONUSED KKKKKKKKKKKKKKK VAMOS LOUD LFG

dont, getting 8th is really good considering the small region they are from!

godyz playing instead today

:))

yeah im down

godyz coming out today

that would be funny ngl

😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹😹

i wish there would be ingame one where we can see speific players pov and shit

my goat

sideshow/sgares/meL

sometimes they are all on the desk so tarik/sliggy/tmv

yes, prx are def going lower bracket

no need, fnatic will 100% win vs loud and fut wil 10000000% win vs prx

math 55 class so hard :(( i need help

prx flairs are always so smart, thank you!!!!!!!

(i) d1(f, f) = R 1
0
|f(x) − f(x)| dx =
R 1
0
0 dx = 0. And, if f 6= g, then
F(x) = |f(x)−g(x)| is not identically zero. Hence ∃x0 so F(x0) = 2 >

0. And by continuity, ∃δ such that ∀x, x0−δ < x < x0+δ, F(x) > . So,
   d1(f, g) = R 1
   0
   F(x) dx =
   R x0−δ
   0
   F(x) dx +
   R x0+δ
   x0−δ
   F(x) dx R 1
   x0+δ
   F(x) dx ≥
   R x0+δ
   x0−δ
   , dx = 2δ > 0
   0 = d1(f, g) = R 1
   0
   |f(x) − g(x)| dx, implies |f(x) − g(x)| = 0 for all x,
   and thus f = g.
   (ii) d1(f, g) is symmetric in definition, as |f − g| = |g − f|, and thus
   equals d1(g, f).
   (iii) d1(f, h)+d1(h, g) = R 1
   0
   |f(x)−h(x)|+|h(x)−g(x)| dx ≥
   R 1
   0
   |f(x)−
   g(x)| dx = d1(f, g).
1. {fn} : ∀ > 0, ∃N(> 1/) so that ∀n > N, x ∈ R we have 0 <
   n
   x2+n2 ≤
   n
   n2 < . Thus, {fn} converges uniformly (and thus also point-wise) to
   0 on all of R.
   {gn} : For any given x, we can pick a sufficiently large N(>
   q
   x2
   
   ) so
   that, n > N implies gn is  close to, 1. Thus, {gn} converges point-wise.
   But, for any n, if we pick x = n to get gn =
   1
   2
   . This means that {gn}
   does not converge uniformly to 1.
2. Let U = {x : d(x, A) < d(x, B)} and V defined anaglgously, with d
   the distance function defined in problem set 2 problem 1. Clearly, U
   and V are disjoint and contain A and B respectivly. Now, it remains
   to show that they are open. Let x ∈ U,  =
   1
   3
   (d(x, B) − d(x, A)), and
   y ∈ B(x). Then, d(y, B) − d(y, A) > d(x, B) − d(x, A) − 2 > 0, and
   we have y ∈ U and U open. So U and V are as desired.
3. I claim that for U, V open, U ⊂ U¯ ⊂ V , there exists W open so that
   U¯ ⊂ W ⊂ W¯ ⊂ V . To see this, consider U¯ and X − V , disjoint closed
   sets. Then there exists W ⊃ U¯, W0 ⊃ X − V open and disjoint. But
   then U¯ ⊂ W ⊂ X − W0 ⊂ X −V . Since X − W0
   is closed and contains
   W, it also contains W¯ . And we have U ⊂ U¯ ⊂ W ⊂ W¯ ⊂ V , as
   desired.
   1
   Now, letting S1 ⊂ X −B, and S0 ⊃ A as produced by X −B and IntA
   with the above lemma applied twice. I inductivly create open Sk/2
   i one
   level of “i” at a time. At each point if q < r, then S¯
   q ⊂ Sr. Sk/2
   i (k
   odd), is generated by the above lemma between S(k−1)/2
   i and S(k+1)/2
   i .
   So of course, all the sets contain the closure of S0 and are contained in
   S1.
   I define f(x) = inf({1} ∪ {r : x ∈ Sr}). This is set is bounded below
   and non empty, so the inf exists. It is clear that f(x) = 0 if x ∈ A,
   f(x) = 1 if x ∈ B, and has range [0, 1]. It now remains to show that
   f is continuous. I first show f(x) = sup({0} ∪ {r : x ∈ X − S¯
   r}). In
   doing this, we only need to consider r in the sets that are terminating
   binary fractions, as for no other r is Sr defined and thus is it possible
   for r to satisfy the condition to be in the sets.
   If, r > f(x) then ∃r0, r > r0 > f(x) with x ∈ Sr0 ⊂ Sr ⊂ S¯
   r, making
   x 6∈ X −S¯
   r. As 0 ≤ f(x), we can upperbound the sup with f(x). And,
   ∀r < f(x) ≤ 1, then ∃r0, r < r0 < f(x) with x 6∈ Sr0 ⊃ S¯
   r and thus
   x ∈ X − S¯
   r. Since all binary fraction 0 < r < f(x) (it is key here that
   f(x) is bounded above 1 so that all of these r produce valid Sr) are in
   the set, and these are dense in the reals, we have f(x) as a lower bound
   for the sup as well. This fails if f(x) = 0, but then, the additional 0
   element saves us. Regardless we have now proved the identity.
   Next, we show f
   −1
   ([0, r)) is open (r is now any real number). I claim
   f
   −1
   ([0, r)) = ∪s<rSs. If x ∈ ∪s<rSs, then ∃s < r such that x ∈ Ss and
   thus, f(x) = inf({1} ∪ {a : x ∈ Sa}) ≤ s < r. If x 6∈ ∪s<rSs, then
   ∀s < r, x 6∈ Ss. Thus, if x ∈ Ss, then s ≥ r; and of course 1 ≥ r.
   So, f(x) = inf({1} ∪ {a : x ∈ Sa}) ≥ r. So x ∈ ∪s<rSs if and only
   if x ∈ f
   −1
   ([0, r)) making the sets equal; as the union of open sets is
   open so is f
   −1
   ([0, r)). The same argument, using the sup definition of f,
   shows that f
   −1
   ((r, 1]) is open. As, f
   −1
   ((a, b)) = f
   −1
   ([0, b))∩f
   −1
   ((a, 1]),
   this set is open as well. Finally, any open set in the reals is the union
   of open balls (one around each point if you like), f
   −1 of any open set
   is the union of open sets and is thus open. Thus, f is continuous as
   desired.
4. Let X be a topological space and F a collection of closed subsets.
   Define G as the complements of sets in F. Then a union of sets in G is
   the complement of an intersection of the corresponding sets in F. So
   2
   F has FIP if and only if G has no finite subcover of X. Also F has the
   total intersection property if and only if G does not cover X. Therefore
   we are done.
5. Let S be a sequentially compact subset of a metric space.
   (i) Suppose S is not closed. Then ∃x 6∈ S, such that ∀r > 0, Br(x)∩S 6=
   ∅. Let, an be a point in S ∩ B1/n(x). Clearly, an → x 6∈ S, and thus
   all subsequences converge to x 6∈ S and thus do not converge in S,
   contradicting the sequential compactness of S.
   (ii) Suppose S is not totally bounded. Then let r be a radius such that
   there is no r net of S. Now define an as follows: Let a0 be any point in
   S and let an ∈ S, such that ∀m < n, d(am, an) ≥ r. Such an an exists
   by the lack of an r net of S. But S sequentially compact implies ∃m, n
   such that d(am, an) < r which is impossible. So, S must be totally
   bounded.
6. Clearly if E is totally bounded, then it is totally bounded relative to
   any metric space containing it. Conversely, if E is totally bounded
   relative to X, then there is some finite set of point p1, p2, . . . pn ∈ X,
   so that every point in E is within /2 of these points. Without loss
   of generality, we may assume that there is some qi ∈ E in each of
   these /2 balls, or else we omit the corresponding pi
   from the original
   enumeration. But now, ∪iB(qi) ⊃ ∪iB/2(pi) ⊃ E, by the triangle
   inequality. And thus for every  we have an epsilon net of E centered
   around points in E, making E totally bounded.
7. Suppose no such r exists. Then for each r, in particular for each 1/n,
   there is some xn such that ∀α, B1/n(xn) 6⊆ Uα. Now, X is compact, so
   exists ni ∈ Z
   +, x ∈ X, so that xni → x. Now, {Uα} covers X, so ∃α, so
   that x ∈ Uα. As these sets are open, ∃r such that Br(x) ⊂ Uα. And,
   by convergence, ∃k > 2
   r
   , so that d(xk, x) <
   r
   2
   . But then, B1/k(xk) ⊂
   Br(x) ⊂ Uα, contradicting our construction of {xn}, and implying the
   existence of such an r

in my calculations it got very closer, but less had the definitive edge.

lmao i dont, i dont give a shit about what other ppl think so upvotes/downvotes mean nothing to me

take ur meds

we said goat, not goat of eco fragging 💀

the greatest player of all time, 2023 and 2024 mvp, back to back champs winner

thats not smthlikeyou11

TenzDeus

im from turkey and my fav team is fut

s0m and marved different playstyles, s0m has better movement and anchoring, marved has better timing and lurks who helped fns manipulate the map. both play the same role but very differently. s0m is a great player.

yay clears ardiis in every way sadly

can someone make an early laugh thread

im from turkey and my fav team is fut

they both play similar roles now that alfa is a sentinel player, alfa isnt playing raze anymore even tho he can

on the sentinel role, with all the evidence gathered - less wins.

didnt talk about the upcoming match

just comparing players

SEN can lose to m80, but does it make nismo better than zekken? (i still love nismo tho)

i paid like 200 lira for it

i am not hating, i am just settling a debate between 2 great players and which one comes up top.

no a single hate message was in this thread

oh shit, my bad kanka

i am never wrong so unfortunately you gonna need to ask someone else for that. try yessirskii?

🙏

i saw this debate happening a lot here, so i decided to give my unbaised and true opinion that is based on real research, stats and trusted top tier analysts opinion.

less clears, your welcome. no need to discuss it further as i have already solved it for you.

i meant mine

i think all fnatic fans are, after all they are facing the 2022 champs

flag and flair sis

im not saying its bad my fellow habibi/kanka

im saying chinese/brazilian aim clears and usually their aim doesnt get them very far like brazilian aim does

as an eu fan, i hope that fnatic/fut wins but i also know theres 0% thats happening because we are such a bad region 😢

turkish aim is overly exaggerated, very fraudulent

nothing that we can do😭😭😭

i also love fnatic, they are so good!!! no shot they lose to loud, loud look so weak and barely could get out of groups

oh shit, i am so sorry fut :(((((( forgive me please :(((