拓扑笔记
Chap 1 Point Set Topology
Lec1
Def: \(T\) topology of X if: (1) \(\empty, X\in T\) (2) ( union of elts in \(T\)) \(\in T\) (3) \(a, b\in T\), then \(a\cap b \in T\).
Notation: elts of \(T\) open; \((X, T)\) topological space; \(C \subset X\) closed if \(X/C\) open; \(A\) neighborhood of \(x\) if \(x \in B \subset A, B\in T\).
Fact: \(A\) open iff \(A\) is nbhd of all its elts.
Lec2
Def: \(f: X\to Y\) continuous if for any \(U\subset Y\) open, \(f^{-1}(U)\) open in \(X\).
Prop: TFAE
(1) \(f\) continuous
(2) \(f^{-1}(C)\) closed in \(X\) for any \(C\) closed in \(Y\)
(3) any \(x\in X\), any nbhd \(V\) of \(f(x)\), \(f^{-1}(V)\) is a nbhd of \(x\)
(Proof: (1) equiv to (2) is trivial. (1) to (3) is trivial. (3) to (1) since the inverse image is nbhd of every elt. )
Example: continuous on standard topology is the usual meaning of continuous.
Def: f homeomorphism if \(f, f^{-1}\) continuous and bijective.
Def: Induced topology \(T_A\) on \(A\sub X\) are the elements \(v\cap A\) with \(v\in T\).
ALS 1
Def. \((X, T_1), (X, T_2)\) topological spaces with \(T_1\subset T_2\), then we call \(T_1\) coarser than \(T_2\), \(T_2\) finer than \(T_1\).
Def. interior \(A^\circ\) are the elements with \(A\) as nhbd
closure \(\overline A\) are the elements \(x\) such that \(X/A\) is not a nbhd of \(x\).
boundary \(\part A=\overline A/A^\circ\)
Prop. \(A^\circ\sub A\sub \overline A\).
Prop. \(A^\circ=\cup_{U\text{ open}, U\sub A}U\), \(\overline A=\cap_{U\text{ closed}, A\sub U}U\).
Prop. \(A^\circ\) is the largest open set \(\sub A\); \(\overline A\) is the smallest closed set contains \(A\).
TFAE: \(A\) open; \(A^\circ=A\); \(A\) is nhbd of every \(x\in A\).
Lec 3
Def. Metric distance \(d\) on \(X\): \(X \times X \to \R^+\), s.t. (1) \(d(x, y)=0\) iff \(x=y\); (2) \(d(x,y)=d(y,x)\); (3) trig ineq;
Let \((X, d)\) be a metric space. Induced topology: open if for each \(x\), a small open ball centered at \(x\) is in the set.
Two metrics on \(X\) are equivalent: within a constant factor.
Conti map between metric spaces: if \(\delta, \epsilon\) language holds.
Lec 4
Basis: \(B\sub T\) a basis if every elt in \(T\) can be written as a union of elts in \(B\).
Subbasis: \(S\sub T\) such that finite intersections of elts in \(S\) is a basis of \(T\).
(Often we only need to check basis. For example: \(f\) continuous, we only need \(f^{-1}(B)\) open for a basis \(B\) of the image.)
Prop: \(B\) is a basis if \(X\) is the union of elts in \(\mathcal B\), and for any \(x\in B_1\cap B_2\), exists \(x\in B\sub B_1\cap B_2\) for \(B\in \mathcal B\).
Cor: \(S\) subbasis only need the union of elts is \(S\).
Product space
Box topology: topology generated by \(\Pi u_i\), with \(u_i\sub X_i\), \(u_i\) open.
Product topology: generated by subbasis \(S = \{\pi^{-1}_j(U_j), U_j\sub_{open} X_j\}\).
Remarks: 1. Fin intersection of \(S\) are those whose all but finite dims are \(X_i\).
- \(I\) finite, then box = product.
- \(\pi_i\) continuous (when the topology on the product space is product topology)
- \(X\) be a space, \(f_i: X\to X_i\) is a map, \(f: X\to \prod x_i, x\to(f_i(x))\), then \(f\) conti iff each \(f_i\) conti.
Lec 6
Quotient topology on \(X/\sim\): \(U\) open if \(\pi^{-1}(U)\) open.
Induced function \(g:X/\sim\to Y\) from \(f: X\to Y\) (need to be well defined)
Then \(g\) surjective iff \(f\) surjective; \(g\) injective if diff class to diff result; \(g\) conti iff \(f\) conti. (\(f=g\pi\), \(g^{-1}=\pi f^{-1}\)).
To show \(g\) Open: suffices to show \(g(u)=f(\pi^{-1}(u))\) open for open \(\pi^{-1}(u)\). (Sometime suffices for basis.)
Lec 7
Def. Nbhd basis of \(x\): a family of nbhds \(B_x\), for any \(x\in U\) open, exists \(B\in B_x, B\sub U\).
Def. Sequence: A map from \(\N\) to \(X\), write as \((x_n)\).
Def. \(X\) a top space, \((x_n)\) a sequence, an element \(x\in X\) is a limit of \((x_n)\) (write as \((x_n) \to x\)) if for any \(x\in U\sub X\) open, exists \(N\ge0\) such that for any \(n > N\), \(x_n\in U\). And say that \((x_n)\) converges to \(x\).
Remark: We only need to check the case \(U\in B_x\).
Examples: in \((X, T_{discrete}), (X, T_{cocountable})\), converge to \(x_n\) iff stationary (all is \(x_n\) after a point. ) (proof: \(X-\{x_n\}+x\) is open)
in \((X, T_{trivial})\), converges to anything
in \((X, T_d)\), nhbd basis are open balls centered at \(x\), so it's the usual understanding of limit.
Def. A topological space \(X\) Hausdoff if for any \(x, y\in X\), then exists disjoint open sets \(U, V\) with \(x\in U, y\in V\).
Prop. Let \(X\) be a Hausdoff space. If \((x_n)\to x, (x_n) \to y\), then \(x=y\). (Proof: ow, exists \(U, V\) Separate \(x, y\). After a point all is in \(U\cap V\), contradicts)
Examples. \((X, T_{discrete}), (X, T_d)\) are Hausdoff, while \((X, T_{trivial})\) is not.
Remark: \(A\sub X\) subspace(induced topology), \(X\) Hausdoff, then \(A\) is Hausdoff.
(In pset4, will prove properties of Hausdoff vs. products and quotients. )
Question: \(X\) Hausdoff implies uniqueness of limits, what about the converse?
Def: \(X\) is first countable if any \(x\in X\) admits a countable nbhd basis. (On Wednesday, we'll see examples and how this is relevant to the question)
Def. A map \(f:X\to Y\) of top. spaces is sequentially continuous if for any \((x_n) \to x\), we have \(f(x_n)\to f(x)\).
Def. \(A\sub X\) sequentially closed if for any \((a_n)\to x\), we have \(x\in A\).
ALS3
SC:
Special cases:
Claim: All subsets of \(X\) endowed with discrete top. are sequentially closed.
Claim: The only SC subsets of \(X\) endowed with triv. top. are \(\empty, X\). (converge to any. )
\((0, 1]\) in \(\R\) with \(\R_{std}\) is not sc. (\(1/n\to 0\))
Prop: Closed subsets are sc. (Ow, \(x_n\to x\) with \(x\in A^c\), take \(x\in U\sub_{open} X\) contradicts).
Map sequentially continuous:
Claim: If \(f:X\to Y\) continuous, then sequentially continuous. (For open \(f(x)\sub U\), \(f^{-1}(U)\) open around \(x\) so finally fall in \(U\). )
First Countability:
Claim: Metrix spaces are first countable. (Simply choose \(B(x, 1/n)\))
Thm: Let \(X\) be a first countable top. space, then
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X Hausdorff iff every convergence sequence converges to a unique point
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\(A\sub X\) closed iff \(A\) sequentially closed
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\(f:X\to Y\) continuous iff \(f\) sequentially continuous.
Lemma: \(X\) first countable, \(x\in X\), there exists a basis of open nghb \(B_x = \{B_n\}\) such that \(B_{n+1}\sub B_n\), and if \((x_n)\) is a seq with \(x_n\in B_n\), then \(x_n\to x\).
Proof: 取nbhd basis的前缀交。
Proof of 1, 2, 3: If not Hausdorff, pick \(B_x(i)\cap B_y(i)\), then two limits.
If not closed, for \(a\notin A\), not open so can pick elts \(B_x(a)\cap A\), then limit \(=a \in A\)
If not continuous, \(f^{-1}(X)\) is not nbhd of \(x\). Pick \(x_n\in B_n(x) - f^{-1}(X)\), finally will converge to \(x\) so \(f\) will converge to \(f(x)\), thus will fall in \(X\).
Lec 8
Def. A topological space \(X\) is connected if for any disjoint \(U,V\sub X\) with \(X=U\cup V\), one has \(U=\empty\) or \(V=\empty\).
Remarks:
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\(X\) not connected if \(X=U\cup V\), \(U, V\) dsjoint, non empty, open in \(X\).
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For \(X, Y\) homeomorphic, then \(X\) connected iff \(Y\) connected.
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\(X\) connected iff the only sets in \(X\) that both open and closed are \(X, \empty\).
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For \(Y\sub X\) subspace, \(Y\) connected iff \(\forall U,V\sub X_{open}\), \(Y\sub U\cup V, Y\cap U\cap V=\empty\) implies \(Y\sub U\) or \(Y\sub V\).
(Negation of both are \(Y=(U\cap Y)\sqcup(V\cap Y)\) for \(U\cap Y, V\cap Y \ne \empty\)
Examples:
(1) \(X,T_{triv}\) connected
(2) If \(|X|\ge 2\), then \((X,T_{discrete})\) not connected
(3) \((0,1)\cup (-1,0)\) in \(\R\) not connected
Prop. Let \(X\) be a space and \(A_i\) connected subspace \(\forall i\in I\). If \(\cap A_i\ne \empty\), then \(\cup A_i\) connected. (Ow, 分成 \(U, V\). 交集中的某个元素在 \(U\) 中则所有 \(A_i\sub U\). )
(Detailed Proof: Ow, assume \(U, V\) disj. with \(\cup A_i\sub U\cup V\) and \(\cup A_i \cap U\cap V=\empty\). Pick \(x\in \cap A_i\), assume \(x\in U\) WLOG. Consider \(A_i\cap U\), \(A_i\cap V\). The second must be empty, so every \(A_i\sub U\). )
Prop. \(A\) space, \(A\sub X\) a connected subspace. If \(B\sub X\) subspace with \(A\sub B\sub \bar A\), then \(B\) connected. (Pf: \(A\) 和开集不交则 \(\bar A\) 也不交)
(Detailed Proof: (Recall: \(\bar A\) is inter of all closed contains \(A\). )
Ow assume open \(U, V\sub X\). Then \((A\cap U), (A\cap V)\) splits \(A\) so can assume \(A\sub U\), \(A\sub X-V\). If exists \(x\in B, x\notin X-V\), then since \(A\sub X-V\), \(x\notin \cap_{A\sub C, C closed} C\), contradicts. )
Prop. \(f\) Continuous, \(X\) connected, then \(f(X)\) connected. (Ow, \(f^{-1}(U)\) and \(f^{-1}(V)\)).
Prop: \(\prod_{1\le i\le n} X_i\) connected iff each \(X_i\) connected.
Proof: Left to right: trivial.
Otherside: induction. Fix \(b\in Y\). Note \(X\times Y=\cup_{x\in X} (\{x\}\times Y)\cup(X\times \{b\})\). Let the term be \(T_x\), then each \(T_x\) connected(two things homeo to \(X,Y\)), and all \(T_x\) inter nonem.
Lec 9
To show: \((a, b) \in \R\) connected.
Proof: Otherwise, assume not inter union of \(U, V\), pick \(u\in U, v\in V\) with \(u<v\) WLOG. Let \(S=\{s\mid [u,s] \in U\}\). \(S\) Non-empty and bounded, so exists \(s_0=\sup S\).
First, \(u\le s_0\le v\), thus \(s_0\in (a, b)\). \(s_0 \notin U\) (or will be larger \(s_0\)), \(s_0 \notin V\) (or will be smaller \(s_0\)).
Cor. \(\R, [a,b],(a,b],[a,b), (-\infty, a), (a, \infty), (-\infty, a], [a, \infty)\) connected.
Proof: pset1 shows that all open intervals are homeomorphic thus connected. \((a,b) \sub (a,b]\sub [a, b]=\overline{(a,b)}\), thus \((a,b]\) connected.
Thm: \(X\) connected, \(f:X \to \R\) continuous, if \(f(a) < r < f(b)\), then exists \(c\in X\) with \(f(c)=r\). (Pf: image of connected should be connected)
(Detailed Proof: Ow, \(f(X)\) not connected (as \(f(X)=(f(X)\cap (-\infty, r))\cup (f(X) \cap (r, \infty)))\). Since \(f\) continuous, so \(x\) not connected. )
Def. Let \(X\) be a top space, a path in \(X\) from \(x\in X\) to \(y\in X\) is a continuous function \(f: [a, b] \to X\) with \(f(a)=x, f(b)=y\).
Def. \(X\) path-connected if \(\forall x, y\in X\), \(x, y\) can be joined by a path.
Rmk: This is a topological notion, so for \(X, Y\) homeomorphic, \(X\) path connected iff \(Y\) path connected.
Prop: If \(X\) is path-connected, then is connected. (Converse is not true. )
Proof: If not connected, is disjoint union of \(U, V\). Pick \(u\in U, v\in V\), \(\gamma: [a, b] \to X\), \(\gamma(a)=u, \gamma(b)=v\). Then \([a,b]=\gamma^{-1}(U)\sqcup \gamma^{-1}(V)\), both open and non empty, so \([a, b]\) not connected, contradicts.
Examples: active learning: \(S^n\) path-connected for \(n\ge 1\).
pet5: \(X\) (Path) connected \(\to\) \(X/\sim\) (path) connected.
Lec 10 & ALS
Def: \(A\sub R^n\) convex: if for any \(x, y\in A\), for any \(0\le t\le 1\), \(tx+(1-t)y\in A\).
Prop: Convex \(\to\) Path-connected. (\(f(t) = tx+(1-t)y\)).
\(\R^n - \{0\}\) path-connected for \(n\ge 2\). (Pf: for \(xy\) not pass through \(0\), easy. Pass through \(0\): pick another \(z\) not on the line. )
Goal: \(\R^n-\{0\}\simeq S^{n-1}\R_{>0} (n>1)\)
Pf: \(f:x\to (x/|x|, |x|)\), has inverse \(g(a, b)\) = \(ba\). Easy to see \(x/|x|, |x|\) continuous as function to \(S^{n-1}, \R_{>0}\), restricting image maintains continuity. \(g\) continous as function from \(\R^n\times \R \to \R^n\), restricting domain maintains continuity.
Goal2: \(S^n\) path connected for \(n>0\). (Since \(S^{n} \R_{>0}\) connected, easy to construct a path for \(a, b\in S^n\).
Prop. \(\R\) and \(\R^n\) not homeomorphic for \(n>1\). (Ow, consider \(\R-\{0\}\), \(B=\R^n-\{f(0)\}\), one path connected while the other not. )
Def. A space \(X\) is locally path-connected if every \(x\in X\) admits a path-connected nbhd.
Prop: Locally path-connected and connected $\to $ path-connected.
Proof: pick \(x\in X\), let \(U=\{y\mid y\in X, \exists \text{ path from } x \text{ to } y \}\). \(U\) Non-empty, suffices to show \(U\) is both open and closed (then \(U=X\)). Open: any \(y\in U\), \(y\) 的 path connected nbhd \(\sub U\). Closed: need \(U=\bar U=\{z\mid X-U \text{ not nbhd of } z\}\). Let \(z\in \bar U\), exists path connected nbhd, has intersection with \(U\).
Connected but not path connected: \(\{(0, 0)\}\cup \{(t, \sin(1/t)) \mid t\in (0, 1)\}\).
