Topology Flashcards
1
Q
- Show that f: R–>R is continuous in the topology sense if and only if it is continuous in the analysis sense.
A
- Hatcher pg. 3-4 or lec notes
2
Q
- Define the product topology and the metric topology. Show that they agree on R^2=RxR.
A
- Def. Let (Xi) be spaces, then the product topology on Prod(Xi) is the topology that is generated by the basis {Prod(Oi) | Oi open in Xi, Xi !=Oi for finitely many i}. (Alternatively the coarsest topology s.t. p_i^-1(Oi) open for all Oi open in Xi for all i)
Def. Let (X,d) be a metric space, then then the topology { O s.s.o. X | for all x in O there is an eps s.t. B(eps,x) s.s.o. O}.
(Alternatively the topology generated by the basis of all open balls)
This is almost norm equivalence.
“=>”: Let O be an open in metric top. Then for all x in O there is an rx>0 s.t. B(x,rx) s.s.o. O (by def of basis). Each ball is contains the subset Ox:=(x1-rx/3,x1+rx/3)x(x2-rx/3,x2+rx/3). Indexing over all x these open covers are equal O=U B(x,rx)=U Ox and so O is open in product top.
“<=”: Let O be open in product top. Then O=U UixVi, where for each i there is some (x1i,x2i) s.t. Ui=(x1i-r1i,x1i+r1i) and Vi=(x2i-r2i,x2i+r2i), then for r:=min(ri1,ri2) B(r,(x1i,x2i)) s.s.o. UixVi and we see O=U UixVi=U B(r,(x1i,x2i)). So I is in metric top.
3
Q
- a) Let A s.s.o. B s.s.o. X, B open. Show that A is open in B if and only if it is open in X
b) What if B is not open?
A
- a)”=>”: A open in B then A=OnB for some O open in X. Finite intersections of open sets are open.
“<=”: A open in X then A=AnB=OnB (since A s.s.o. B) with O open in X, so A open in B
b) A=B=[0,1], or X=R^2, B=Rx{0}, A=(0,1)x{0}
4
Q
- Define connected and path-connected. Show that a path-connected space is connected.
A
- Def. (X,T) top space is connceted if it is not disconnected, i.e. if (AT LEAST) ONE of the following conditions is NOT satisfied:
i) Ex O1,O2 disjoint and open in X s.t. O1uO2=X
ii) Ex C1,C2 disjoint and open in X s.t. C1uC2=X
iii) Ex A clopen in X s.t. A is neither {} nor X.
Def. X is path connected if for every x,y in X there is a path f s.t. f(0)=x, f(1)=y.
Pf: Let O1,O2 be open and disjoint in X s.t. O1uO2=X. Pick any two points, by contradiction one in each set, to get a path f. Then [0,1]=f^-1(X)=f^-1(O1uO2)=f^-1(O1)uf^-1(O2) (where last union is still disjoint and open). Contradiction
5
Q
- Prove that Q is totally disconnected.
A
- Def. Totally disconnected means that C(x)={x} for all x in X.
Suppose there was an x s.t. C(x) is strictly larger that {x}. Then there is a distinct y in C(x). But now C(x)=((-inf,z)nC(x))u(C(x)n(z,inf)) for z in R\Q. x is in one, y in the other and they are disconnected. So C(x) is disconnected, contradiction.
6
Q
- Show with pictures that the Cantor set is totally disconnected.
(Def. isolated point)
A
- (Def. x is isolated point if {x} is open. Cantor set is non-empty, has no isolated points and is totally disconnected)
Set in R is connected iff C is interval.
Recall: C(n+1):=C(n)\Union( ( (3k+1)/3^n,(3k+2)/3^-n) ): k from 1 to 2^n - 1 ) and C:=Intersection(C(n)).
Suppose there is an interval (a,b) in C. Then (a,b) in Cn for all n. But this cannot happen because L(Cn)=(2^n-1)*3^-n=(2/3)^n-1/3^n and for every eps:=b-a there is an n s.t. eps>L(Cn). So (a,b) is contained in a set that is shorter than itself. Contradiction.
7
Q
- Is X=R^2-Q^2 connected?
A
- It is path connected: Consider (x,y),(z,w) in X.
Def. (x,y)-lin->(v,w) to mean the path f(t):=((1-t)x+tv,
(1-t)y+tw) in R^2 ! Let (x,y)-lin->(v,w)-lin->(s,t):=((x,y)-lin->(v,w))*(v,w)-lin->(s,t)).
1) x in R\Q, in which case
a) v in R\Q: (x,y)-lin>(x,pi)-lin->(v,pi)-lin->(v,w)
b) v in Q then w in R\Q: (x,y)-lin->(x,w)-lin->(v,w)
2) x not in R\Q, so y in R\Q, in which case:
a) w in R\Q: (x,y)-lin>(pi,y)-lin->(pi,w)-lin->(v,w)
b) w in Q then v in R\Q: (x,y)-lin->(v,y)-lin->(v,w)
8
Q
- Show the product of path-connected spaces is path-connected.
A
- Take any two points (xi)i,(yi)i Prod(Xi). For each coordinate find a path gamma_i from xi to yi. Then (gamma_i)i is a path in the product space since all its components are continuous. To show it is continuous:
Let O be open in product top , then O=Prod(Oi : finite I)xProd(Xi : i rest) then (gamma_i)i^-1(O)=Intersection(gamma_i^-1(Oi))x[0,1]
9
Q
- a) Are connected components closed? Prove it.
b) What about path-connected coponents?
[ c) When do path-connected components and connected components coincide?]
A
- Yes.
Lemma: X top space, A s.s.o. X connected, then bar(A) remains connected.
Pf: Suppose not and let bar(A)=C1 u C2 with Ci disjoint and clopen. Then A=(AnC1) u (AuC2), where AnCi clopen i A (check). Therefore one of them is empty set (for A is connected). W.l.o.g. AnC1={}. Then A s.s.o. C2. Since C2 is closed and bar(A) is by a def the smallest closed set containing A, it must be that bar(A) s.s.o. C2, i.e. bar(A)=C2, hence C1={} and the only clopen sets are bar(A) and {}. So bar(A) is connected.///
a) By lemma bar(C(x)) is connected. But since C(x) is the largest w.r.t. inclusion C(x)=bar(C(x)) and so C(x) is closed.
b) No. Topologists sine curve: G={(x,sin(1/x)) : x>0} is path-connected, but its closure Z=Gu{(0,x) : x in [-1,1]} is not path-connected, so that it is not equal to its closure and therefore not closed.
[c) Def. X is locally path connected if for every x there is an open path connected ngbh of x.
Fact 1: Locally path connected => C(x)=P(x)
Fact 2: Connected and open in R^n => p.c. and C(x)=P(x)]
10
Q
- Let {C(n)} be connected subspaces of X. Suppose that C(n) n C(n+1) != {} for all n. Show that U C(n) is connected.
A
- Let K1, K2 be disjoint and clopen s.t. U C(n) = K1 u K2 (these exist for one could choose e,g, K1={}). Then C(n)=(K1 n C(n) ) u (K2 n C(n) ) but C(n) connected so C(n) s.s.o. K1 or s.s.o. K2 for every n.
W.l.o.g. let C(0) s.s.o. K1. Suppose C(n) s.s.o. K1 for arbitrary n, then since C(n) n C(n+1) != {} also C(n+1) s.s.o. K1. Hence, C(n) s.s.o. K1 for all n, so K1=U C(n) and K2={} and U C(n) is connected.
11
Q
- Let f:S^1–>R be continuous. Show that there exists (x,y) in S^1 so that f((-x,-y)).
A
- Def. g(t):=(cos(t),sin(t)), note g(t+pi)= -g(t).
Def. also phi(t):=f(g(t))-f(g(t+pi)).
Now for phi(0)= -phi(pi). So by the intermediate value thm, there is a t* s.t. phi(t)=0. Then for z:=g(t), it holds that f(z)=f(-z).
Lem: (intermediate value thm) f:[a,b]–>R and for each c s.t. f(a)c or f(y)
12
Q
- Prove that for any countable set A s.s.o. R^2, we have that R^2\A is connected.
A
- Yes, it is even path connected. Let x,y be in R^2\A. Consider their respective hedgehog spaces in R^2, i.e. all lines intersecting x and y. Remove from these spaces all spines that intersect with A and the two lines, one in each space, that are parallel to each other. We still have uncountably lines to choose from. We only need one each. These intersect somewhere, call this point z. We can now construct the path x -lin-> z -lin-> y.
13
Q
- Prove that the Cantor space X={0,1}^N is homeomorphic to XxX, where N=natural numbers.
A
- Let x=(x(1),x(2),…) in X and y=((y(0,1),y(1,1)),(y(0,2),y(1,2)),…) XxX, where x(i),y(0,i),y(1,i) in {0,1}.
Take f:X—>XxX, with fi(x(i)):=y(i mod 2+1,roundup(i/2)). Its inverse is f^-1(y(k,i)):=x(2(i-1)+k+1).
It’s easy to check that these are inverses of each other. Both are continuous, since we are using the discrete topology on {0,1} and so the product topology, the box topology and the discrete topology coincide.
14
Q
- Let X be a metric space and A,B disjoint closed s.s.o. of X. Show that there exists a continuous function f:X–>[0,1] with f(A)={0} and f(B)={1}.
A
- Define f(x)=d(x,A)/(d(x,A)+d(x,B)), where d(x,C):=inf{d(x,c):c in C}=min (since C closed) and this is well defined as the denominator is never 0, since sets are disjoint.
This is 0 on A and 1 on B. Function is always in [0,1].
It is continuous as a continuous combination of continuous functions.
Metric is continuous:
https://math.stackexchange.com/questions/528220/metric-is-continuous-on-the-right-track
15
Q
- Prove that there is no injective path connecting the two zeros in the line with two zeros.
A
- Let f be the path from 01 to 02. By injectivity f(1/2)=c is neither of the zeros. The image of the path restricted to [0,1/2] and [1/2,1] respectively by the intermediate value theorem both cross c/2. So the restrictions give us t0 and t1. But then also the unrestricted path maps both t0 and t1 to c/2, which contradicts injectivity.
16
Q
- Show that the line with two zeros is path-connected.
A
- Let x,y be in Ru{01,02}=Rx{0,1}/{(x,0)~(x,1) iff x!=0}, for ease of notation we will work on first set.
If neither x nor y is in {01,02} then we can concatenate two linear paths ending and starting at a favorite 0.
[define f on [0,1] piecewise as f(t):=(1-2t)x if t in [0,1/2), f(1/2):=0i, f(t):=(2t-1)y.]
If only one is a 0, say x=01, then define f(0):=01=x and f(t):=ty for t in (0,1].
If both the are the same zero, we wait.
If each is a different zero, say x=01 and y=02, then define piecewise f(0):=01=x, f(t):=t for t in (0,1/2], f(t):=1-t for t in [1/2,1) and f(1):=02.
17
Q
- Prove that the cofinite topology on R is not Hausdorff.
A
- Let Ox and Oy be open neighborhoods of x and y. Then R\Ox and R\Oy are finite, and so R\Ox u R\Oy=R(Ox n Oy) is finite. The complement of a finite set in R cannot be finite, so Ox n Oy is not finite, let alone empty!
Suppose it were Hausdorff. Let x,y in R. Then we get Ox,Oy in in {R\F : F finite} u {{}} s.t. Ox n Oy = {}, which is finite. Since R\Ox and R\Ox is finite, so is R\Ox u R\Oy=R(Ox n Oy)=R but this is absurd.
18
Q
- Give an example of a non-Hausdorff space X with a point p so that X{p} is Hausdorff.
A
18.
+R with two zeros. If we remove one, we get R, which is a metric space and as such Hausdorff.
+Two points with the trivial topology.
+ Niki says: take any Hausdorff space X and remove a point p from every set (besides X) in the topology besides X or add a point p in every set (besides {}) in the topology.
19
Q
- Prove that for all pairs of distinct points {p,q} in S^2, S^2{p,q} is homeomorphic to S^2{(0,0,1),(0,0,-1)}.
A
- If p=(0,0,1),q=(0,0,-1) done. So suppose not.
We know that S^2 is homeomorphic to the one point compactification of R^2 or C i.e. to S^2~Cu{inf}. We have the map (stereographic projection) H s.t. S^2{p,q}~(Cu{inf}){H(p),H(q)}.
Use the map f:(C u {inf}){p,q}–>C{0} with f(z):=(z-p)/(z-q) with inverse f^-1(z)=(qz-p)/(z-1) (both are continuous Möbiustransformations).
Now S^2{p,q}~(Cu{inf}){H(p),H(p)}~C{0}~S^1{(0,0,-1)}, where in the last step we use the typical stereographic projection.
Def. Let X be a non-cpt Hausdorff top space. The one-point compactification X^ of X is Xu{inf} with the topology (to be proved): {U s.s.o. X : U open} u {X\C u {inf} : C cpt}.
20
Q
- Let f:X–>Y be continuous, and let Y be Hausdorff. Show that G:={(x,f(x))} s.s.o. XxY (the graph of the function) is closed.
A
- WTS: XxY\G open.
Let (x,y) in XxY\G, then f(x)!=y. Because Y is Hausdorff, there exist disjoint open neighborhoods Of(x), Oy s.s.o. Y. By continuity of f, O:=f^-1(Of(x)) is an open neighborhood of x. Then OxOy open neighborhood of (x,y).
Claim: OxOy s.s.o. XxY\G
Pf: Let (x’,y’) be in OxOy. Then x’ in f^-1(Of(x)) and y’ in Oy.
By the former f(x’) in Of(x). But since Of(x) n Oy ={} by construction f(x’)!=y’ and so (x’,y’) not in G.
21
Q
- a) Give an example of a continuous bijection which is not a homeomorphism.
b) Can you give a general criterion for showing that a continuous bijection is a homeomorphism?
A
- a) Let X set with |X|>1 and T1:=discrete topology and T2:=trivial topology, then any bijection f:(X,T1)–>(X,T2) is continuous but not bicontinuous i.e. a homeomorphism. For let O in T2 be neither the empy set nor X, then f^-1(O)=O is not in T1.
We could also set e.g. X=[0,1] and T1=std top. Test U=(0,0.5) and see.
b) This property is called openness. f:X–>Y is open if for every open set O in X, f(O) is open in Y.
If f is open, continuous and bijective then it is a homeomorphism by definition.
[Also we showed that if X is cpt. and Y is Hausdorff then every continuous bijection f:X–>Y is closed and a homeomorphism]
22
Q
- a) Define the limit of a sequence.
b) Prove that the limit of a sequence is unique if X is Hausdorff.
A
- a) Def. Let X top space and a sequence (xn)n in X. Then x is called the limit of (xn)n if for every open O s.s.o. X, it holds that |{n : xn not in O}|Nx: xn is in Ox
and there is an Ny s.t. for all n>Ny: xn is in Oy,
so for all n>max(Nx,Ny): xn is in Ox n Oy={}. Contradiction!
23
Q
- a) Prove that a subspace of a Hausdorff space is Hausdorff.
b) What about normal?
A
- a) Let x,y in Y. Then x,y in X and so they have open disjoint nbhs Ox and Oy respectively. Then Ox n Y and Oy n Y are also disjoint and open in Y by def of subspace top.
b) Def. X is normal if {x} is closed for all x in X and all disjoint closed subsets C1, C2 of X there are two disjoint and open supersets.
[0,1] cpt so [0,1]^R cpt too by Tychonoff’s thm.
[0,1] HD, so [0,1]^R HD too (see below). By a Prop. from the lecture, we know that if X is cpt and HD, then X is normal. So, [0,1]^R is normal.
Now (0,1)^R~R^R is a subspace but it is not normal…..[Pf..?]
Lem: Xi Hd => Prod(Xi) Hd
Pf: Let x,y in Prod(Xi) s.t. x!=y. Then there is a i s.t. xi!=yi. Then there is a Oxi and Oyi disjoint open neighborhoods in the Hausdorff space Xi. pi^-1(Oxi) and pi^-1(Oyi) are therefore disjoint open neighborhoods of x and y. So Prod(Xi) is Hd.
Another two examples are the Sorgenfrey plane (product space or R with half open intervals [a,b) ) and the Moore plane (closed top half plane with open balls lying within top half plane or tangent to x-axis)
24
Q
- Let X be first-countable. How does one characterise the closure of some A s.s.o. X in terms of limits?
(Def. 1st countable)
A
- Def. X is first-countable if each point x in X has a countable neighborhoods basis, i.e. a family of open neighborhoods {On} s.t. for all open neighborhoods O of x, there is an n s.t. On s.s.o. O.
Characterisation in X f.c.: bar(A)={x: x is a limit of a sequence in A}
Pf: “=>”: If a in bar(A) then there is a countable neighborhood basis of a, that can be chosen to be decreasing (by O’n:=On n On-1), for a in bar(A) and any open neighborhood, OnA is non-empty. Now for every n choose a xn in On n A != {}.
“<=”: If x is a limit of (xn)n s.s.o. A. Then for every O there {n:xn not in O} is finite, in particular its complement is infinite and hence O contains a xn, which is also in A, so OnA!={}.
25
Q
- Prove that the cofinite topology on R is not first countable.
A
25. + Let x in R.
\+ {On} countable nbh. basis of x.
\+ Fn:=R\On finite.
\+ F:={x} u Union(Fn) countable.
\+ Let y in R\F (non-empty)
\+ R\{y} open and x in R\{y}, so On s.s.o. R\{y} for some n
\+ y not in On, i.e. in Fn
\+ y in F. Contradiction!!
26
Q
- State and prove the characterisation of continuity in terms of limits for first-countable spaces.
A
- Char: X first countable. f:X–>Y contin iff (x lim of (xn)n => f(x) lim of (f(xn))n.
“=>”: Let (xn)n have limit x. Let Of(x) be an open nbh of f(x). Then Ox:=f^-1(Of(x)) is an open nbh of x. Now since xn in f^-1(O) implies f(xn) in O by counterposition f(xn) not in O implies xn not in f^-1(O). So {n : f(xn) not in O} s.s.o. {n : xn not in f^-1(O)} the latter of which is finite.
“<=”: Suppose f is not continuous. Then there is an open O s.s.o. Y s.t. f^-1(O) s.s.o. X is not open. Then f^-1(O) != {}. So let x be in f^-1(O) s.t. for every open nbh Ox of x, Ox n f^-1(O)^c != {} (if such an x would not exist then f^-1(O) would its interior and open). Let {On}n be a nbh basis of x. By the above we can choose an xn in On n f^-1(O)^c. This defines a sequence converging in x. However, f(xn) not in O for all n, i.e. |{n : f(xn) not in O}|=inf. So f(xn) does not converge in f(x).
27
Q
- Let X be first countable, and let (xn)n be a sequence. Also, let x be so that every neighborhood of x contains infinitely many points of the sequence. Show that a subsequence of (xn)n converges to x.
A
- Let (On)n be a non increasing neighborhood basis of x. Choose xn0 in O0. Define O’0:=O0 and O’k:=Onk\ Union( {xi} for i in {0,1,…,nk-1}) and pick xnk in O’k. This is well defined as O’k still contains infinitely many sequence points. This converges towards x by construction. [The reason we remove points is to really make the new sequence a subsequence.]
28
Q
- Show that metric spaces are Hausdorff.
A
- Let x,y be in X. Choose r s.t. d(x,y)/2>r>0. Then B(r,x) and B(r,y) are disjoint and open.
29
Q
- Prove that R is not homeomorphic to R^2.
A
- The keyword is cutpoints for their number are topological invariant. R has infinitely many and R^2 has none.
+To show it in this case, suppose they were homeomorphic. Then R{0}~R^2{H(0)} too. But the former has fundamental group Z (deformation retracts onto S^1 by homotopy: F(x,t):=(1-t)x+tx/|x|) but R^2{H(0)} has trivial fundamental group. So they are not homotopic and therefore not homeomorphic.
- The keyword is cutpoints for their number are topological invariant. R has infinitely many and R^2 has none.
30
Q
- When is the discrete topology compact?
A
- It is compact iff it is finite.
“=>”: Let X=Union(Oi), then since X is discrete X=Union(Union({xj} : xi in Oi):i in I)=:Union(xij : (i,j) in IxJ). Since X compact there is a finite subcover: X={x1}u…u{xn}, i.e. X is finite.
“<=”: If X is finite, then every open cover is its own finite subcover.
31
Q
- Prove that closed intervals are compact.
A
- Hatcher pg. 31
32
Q
- Let X be a topological space and Y, Z subsets of X.
Suppose that Y and Z are compact, and that X=union(Y,Z).
Show that X is compact.
A
- Take an arbitrary cover of X. Since Y and Z are in X, this cover also covers Y and Z. Since they are compact, we know that they both posses finite subcovers for this open cover. Taking the union of these two subcovers proves the claim.
33
Q
- Show that if X is compact and C is a closed subset of X, then C is compact.
A
- Let (Oi)i be a cover of C. Then (Oi)i u {C^c} is a cover of X. Hence there is a finite subcover {O1,…,On} u {C^c}. Hence {O1^c,…,On} is an open cover of C.
34
Q
- State the characterisation of compactness for metric spaces. Prove one of the implications.
A
- L10: Photo 20.03.19, 14 53 52
35
Q
- Give at least two proofs that IR is not compact.
A
- a) Def. On:=(-n,n) then (On)n is an open cover of IR. So there is a finite subcover. But set has a maximum and minimum. Contradiction. (Alternative cover: Om:=(m-c,m+c), c>1/2, m in Z.
b) By Heine-Borel every compact subset of IR is closed and bounded. But IR is not bounded.
c) IR isn’t totally bounded like compact metric spaces should be. (char (metric space): cpt iff t.b. & complete).
36
Q
- Show that a continuous function f:X–>IR, where X is compact, has a maximum.
A
- X cpt and f contin, so f(X) is compact (provable by open cover argument). Since f(X) s.s.o. IR it is closed and bounded. Since it is bounded the supremum (and infimum) exists in IR, and since it is closed, the supremum is a maximum (and infimum is a minimum).
37
Q
- Let X be compact, Y Hausdorff, f:X–>Y is continuous. Show f is closed i.e. if C is closed then so is f(C).
A
37. C closed subset of the compact space X, so C is also compact [Q33: prove, for 4+] So f(C) is compact [prove for 4+] in a Hausdorff space and hence closed [prove for 4+, Source: Photo 01.04.19, 09 27 35, (Hatcher pg. 35) or https://math.stackexchange.com/questions/83355/how-to-prove-that-a-compact-set-in-a-hausdorff-topological-space-is-closed].
38
Q
- Let X be compact and F.X–>Y be continuous and surjective. Prove that Y is also compact.
A
- Let (Oi)i be a cover of Y. Then (f^-1(Oi))i is a open cover of X be continuity. So there is a finite subcover {f^-1(O1),…,f^-1(On)}. Since f is surjective {O1,…,On} is a finite subcover of Y.
39
Q
- Prove that a compact Hausdorff space is normal.
A
- [L14: Photo 03.04.19, 13 22 07] uses
Lemma: X cpt. & HD => X normal [L13: Photo 01.04.19, 09 27 54]
Pf lem: Let B be closed. Then it is open. For a (in A) and b in B we get disjoint open neighborhoodss Uab, Vab. Consider the cover {Vab}_b of the B with a finite subcover (Vab1,…,VabN with corresponding Uab1,…,UabN). Now the intersection of the Uab1,…,UabN is also open (nbh of x) and does not intersect B. Hence X is normal. q.e.d.
Pf of statement: For every a in A we get open neighborhood Oa and open superset OBa of B (by lem). So (Oa)a defines a cover of A with finite subcover {Oa1,…,Oan} with associated {OBa1,…,OBan}. Let U be union of first finite collection and V the intersection of the second. Now U and V are disjoint open supersets of A and B.
40
Q
- Let X be a compact metric space. Suppose f:X–>X is continuous and f(x) != x for all x in X. Prove that there exists a epsiolon>0 s.t. d(x,f(x))>epsilon for all x in X.
A
- XxX is compact by Tychonoff, d is continuous on compact XxX, so d takes on minimum m>/=0 but since f(x) != x, d(x,f(x)) != 0, so m>0. There is an epsilon s.t. m>epsilon>0. So d(x,f(x))>epsilon for all x in X.
Why doesn’t the following work:
Suppose there was a x0 in X s.t. d(x,f(x))=epsilon for all epsilon. Then d(x,f(x))=0 and f(x)=x. Contradiction???
41
Q
- Let (xn)n be a sequence in a topological space X with limit x. Show that {xn}u{x} is compact.
A
- Let (Oi)i be an open cover of {xn}u{x}. Then there must be a i s.t. x in Oi=:O. By def of limit (|{n|xn in O}|
42
Q
- Let (X,d) be a compact metric space. Show that sup{d(x,y)|x,y in X} is achieved.
A
- a) Since X cpt. XxX is also by Tychonoff. d is continuous [show!!! for 3+] on cpt. space, so d takes on its maximum=supremum (by Q36).
b) Since X is cpt and metric, so is XxX [Tychonoff, induced p-norm/sup norm]. Since d continuous [show!! for 3+], IR is Hausdorff, d is a closed map [show!!]. This implies that d(XxX) stricly in IR closed and therefore has maximum.
Lemma: d:XxX–>IR is continuous
Pf: https://proofwiki.org/wiki/Metric_is_Continuous
43
Q
- Define the quotient topology. And prove it is indeed a topology.
A
- [L14: Photo 03.04.19, 14 18 34]
Prop/Def: X top space, Y set, f:X–>Y (surjective?) function then {U s.s.o. Y | f^-1(U) open s.s.o. X} is a topology on Y called the quotient topology.
Pf: *f^-1({ })={ } open in X => { } open in Y
*f^-1(Y)=X open in X => Y open in Y
*Supp. Oi open in Y for all i => f^-1(Oi) open in X for all i
=> f^-1(Ui(Oi))=Ui f^-1(Oi) open in X => Ui(Oi) open in Y
*Supp. O1,…,On open in Y => f^-1(O1),…,f^-1(On) open in X
=> f^-1(O1n…n On)=f^-1(O1) n…n f^-1(On) open in X => O1n…n On open in Y
44
Q
- Prove that a quotient of a connected space is connected.
A
- Suppose Y not connected. Then there are two open disjoint non-empty sets O1,O2 in Y s.t. Y=O1 U* O2 (O1Y=O1 U O2 & O1nO2={ }). Then X=p^-1(Y)=p^-1(O1 U* O2)=p^-1(O1) U* p^-1(O2) [disjointness: for if not then element in intersection gets mapped into two disjoint set. Contr.]. So X is not connected.
45
Q
- Prove that the Klein bottle is Hausdorff.
A
- Def. of Klein bottle. Mention it can be enbedded in R^4 but as far as I know requires some differential geometry. Mention it is homeomorphic to Torus quotient but don’t fully understand. So show homeomorphic to union of two Möbius strips?
I think following is wrong:
Photo 10.04.19,14 48 45
Prop: X Haus, ~ Equiv rel.
R={(x,y)|x~y} s.s.o. XxX.
Suppose p:X-->X/~ is open,
then X/~ is Hausdorff iff R is closed.
Pf closed #1: My idea: R={((t,0),(t,1))|t in [0,1]}u{((t,1),(t,0))|t in [0,1]}u{((x,y),(x,y))|x,y in [0,1]}u{((0,t),(1,1-t))|t in [0,1]}u{((1,1-t),(0,t))|t in [0,1]}=:R1u...uR4. These sets are all compact/closed in XxX "s.s.o." IR^4 as one can see after rewriting.
Pf closed #2: Kirill: Show that each of the above subsets is the complement of an open set e.g. ((t,0),(t',1)) for t!=t' is in complement of R1.
46
Q
- Let p:X–>Y be a quotient map, with Y connected and p^-1({y}) connected for each y in Y. Prove that X is connected.
A
- Def. A quotient map is a map f:X–>Y that is surjective and s.t. O s.s.o. Y open iff f^-1(O) s.s.o. X open.
Let X=O1 u O2 be a partitioning of X into two open sets. Define Y1:={y in Y : p^-1({y}) s.s.o. O1} and Y2:={y in Y : p^-1({y}) s.s.o. O2}.
+Now for every y in Y either p^-1({y}) s.s.o. O1 or s.s.o. O2 but not both since p surjective (so p^-1({y}) non-empty) and p^-1({y}) connected. So Y=Y1uY2 disjoint and so p^-1(Yi)=Oi.
+Y1 and Y2 are therefore open by def of quotient map.
+So Y1 or Y2 must be empty, since Y is connected. Hence O1 or O2 must be empty. So X is connected.
