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Topology problem sheet 3

 
  • Jan 16, 2023

Compactness

Let $\def\d{\mathrm{d}}\d$ be one of the metrics $\d_1,\d_2,\d_∞$ on $ℝ^n$. Let $K$ be a non-empty compact subset of $ℝ^n$, and let $F$ be a non-empty closed subset of $ℝ^n$. We define $$\DeclareMathOperator{\dist}{dist} \dist(K, F)=\inf\{\d(x, y): x∈K,y∈F\}.$$ Prove that there exists $a∈K$ and $b∈F$ such that $\d(a, b)=\dist(K, F)$.
There is a sequence $(a_n,b_n)∈KΓ—F,\d(a_n,b_n)<\dist(K,F)+\frac1n$. Since $K$ is compact, there is a subsequence $a_{n_k}$ converging to some $a∈K$ with $\d(a_{n_k},a)<\frac1{n_k}$.
Let $F_k=B(a,\dist(K,F)+\frac2{n_k})∩F$. Then $F_kβŠ‚B(a,\dist(K,F)+\frac2{n_k})$ and $F_k$ is closed, [A closed subset of compact set is compact] $F_k$ is compact.
By triangle inequality, $\d(a,b_{n_k})<\d(a_{n_k},a)+\d(a_{n_k},b_{n_k})<\dist(K,F)+\frac2{n_k}$, so $b_{n_k}∈F_k$.
$F_k$ is a sequence of nested non-empty compact sets, so $βˆƒbβˆˆβ‹‚_kF_k$, but $F_kβŠ†F$, so $b∈F$, so $\d(a,b)β‰₯\dist(K,F)$.
$βˆ€k:b∈F_kβ‡’βˆ€k:\d(a,b)<\dist(K,F)+\frac2{n_k}β‡’\d(a,b)≀\dist(K,F)β‡’\d(a,b)=\dist(K,F)$
Let $(X,𝒯)$ be a topological space and let $C=Xβˆͺ\{∞\}$ where $∞$ denotes some extra point not in $X$. Let $𝒯'$ denote the union of $𝒯$ with all subsets of $C$ of the form $Vβˆͺ\{∞\}$ where $VβŠ†X$ and $Xβˆ–V$ is compact and closed in $X$.
Prove that $(C,𝒯')$ is a compact topological space containing $(X,𝒯)$ as a subspace.
This is called the one-point (or the Alexandrov) compactification of $X$. Prove that the one-point compactification of $ℝ^2$ is homeomorphic to the 2-sphere.
General Topology by Jacques Dixmier, page 47 Β§4.5.8
$βˆ…βˆˆπ’―βŠ‚π’―',C=(Xβˆ–βˆ…)βˆͺ\{∞\}βˆˆπ’―'.$
For $U_iβˆˆπ’―'$, we prove $U≔⋃_{i∈I}U_iβˆˆπ’―'$.
$I≕JβˆͺK$ where $βˆ€i∈J:U_iβˆˆπ’―;βˆ€i∈K:U_i=(Xβˆ–V_i)βˆͺ\{∞\}$, $V_i$ is closed and compact in $X$, so $V=β‹‚_{i∈K}V_i$ is compact in $X$. If $K=βˆ…$ then $Uβˆˆπ’―βŠ‚π’―'$. Suppose $Kβ‰ βˆ…$, then $∞∈U$. Also $Xβˆ–U=Vβˆ©β‹‚_{j∈J}(Xβˆ–U_j)$ is a closed subset of $V$ so it is compact in $X$, thus compact in $C$. Hence $Uβˆˆπ’―'$.
For $U_1,U_2βˆˆπ’―'$, we prove $U_1∩U_2βˆˆπ’―'$.
If $βˆžβˆ‰U_1,U_2$ then $U_1,U_2βˆˆπ’―β‡’U_1∩U_2βˆˆπ’―βŠ‚π’―'$.
If $∞∈U_1,βˆžβˆ‰U_2$ then $βˆžβˆ‰U_1∩U_2$ and $U_1βˆ–\{∞\},U_2βˆˆπ’―$, so $U_1∩U_2=(U_1βˆ–\{∞\})∩U_2βˆˆπ’―βŠ‚π’―'$.
If $∞∈U_1,U_2$ then $∞∈U_1∩U_2$ and $Xβˆ–U_1,Xβˆ–U_2$ are compact$β‡’(Xβˆ–U_1)βˆͺ(Xβˆ–U_2)=Xβˆ–(U_1∩U_2)$ is compact$β‡’U_1∩U_2βˆˆπ’―'$.
So $(C,𝒯')$ is a topological space, we prove it is compact: Let $\left(U_i\right)_{i∈I}$ be an open covering of $C$. There exists $i_0∈I$ such that $∞∈U_{i_0}$. Then $$ U_{i_0}=(X-G)βˆͺ\{∞\}, $$ where $G$ is compact. The $U_i$ cover $G$, therefore there exists a finite subset $J$ of $I$ such that $\left(U_i\right)_{i∈J}$ covers $G$ Then$$C=U_{i_0}βˆͺ\left(\bigcup_{i∈J} U_i\right) . $$The steorographic projection is a homeomorphism from the 2-sphere to the one-point compactification of $ℝ^2$.
Let $l^∞$ be the space of bounded real sequences, with the norm ${‖𝐱‖}_∞=\sup _{nβˆˆβ„•}\left|x_n\right|$, where $𝐱=\left(x_n\right)_{nβˆˆβ„•}$.
Prove that the closed unit ball of $l^∞$, $B'(𝟎, 1)=\set{β€–π±β€–βˆˆl^∞:{‖𝐱‖}_βˆžβ‰€1}$, is not compact.
Thus, $B'(𝟎, 1)$ is a closed bounded subset in a complete normed vector space which is not compact.
Let the $n$th term of $𝐱_n$ be 1 and others be 0. For any $n$, ${‖𝐱_nβ€–}=1⇒𝐱_n∈B'(𝟎, 1)$. Also ${‖𝐱_i-𝐱_jβ€–}=1$ for any $i,j$. So the sequence $(𝐱_n)$ doesn't contain Cauchy sequence. So $B'(𝟎, 1)$ is not sequentially compact, thus not compact.

Quotient spaces

The integer part of a real number $x$ is the unique integer $nβˆˆβ„€$ such that $n≀x<n+1$. Denote it by $I(x)$.
On ℝ we define the relation $xβ„›y⇔I(x)=I(y)$.
  1. Prove that β„› is an equivalence relation.
  2. Let $p: ℝ→ℝ/β„›$ be the quotient map, let $ℝ/β„›$ be endowed with the quotient topology, and let $U$ be an open set in $ℝ/β„›$. Prove that if $nβˆˆβ„€$ is such that $p(n)∈U$ then $p(n-1)∈U$.
  3. Deduce that the open sets in $ℝ/β„›$ are βˆ…, $ℝ/β„›$ and the image sets $p(-∞, n]$, where $nβˆˆβ„€$.
  4. Is the map $I: ℝ→℀, x↦I(x)$ continuous (when β„€ is endowed with the subspace topology)?
    Prove that $I$ defines a bijection $\tilde{I}: ℝ/β„›β†’β„€$. What is the topology on β„€ making $\tilde{I}$ a homeomorphism?
  1. $I(x)=I(x)β‡’xβ„›x$
    $xℛy⇒I(x)=I(y)⇒I(y)=I(x)⇒yℛx$
    $xℛy,yℛz⇒I(x)=I(y)=I(z)⇒xℛz$
  2. $UβŠ‚β„/β„›$ is open, by definition of quotient topology, $p^{-1}(U)βŠ‚β„$ is open, but $n∈p^{-1}(U)$,
    so $βˆƒΞ΅<1:B(n,2Ξ΅)βŠ‚p^{-1}(U)β‡’n-Ρ∈p^{-1}(U)β‡’p(n-Ξ΅)=p(n-1)∈U$.
  3. βˆ… is open by definition. If $UβŠ‚β„/β„›$ is open, $p(n)∈U$, by (b) $p[n-1,n)βŠ‚U$, by induction $p(-∞,n]βŠ‚U$.
    Therefore $U=ℝ/β„›$ if $\sup p^{-1}U=∞$; $U=p(-∞, n]$ if $\sup p^{-1} U=n$.
    On the other hand, $p^{-1}(-∞,n]=(-∞,n+1)$ is open, so $(-∞,n]$ is indeed open in ℝ/β„›.
  4. The subspace topology on β„€ is discrete, I isn't constant, so not continuous. Direct proof: $I^{-1}(\{0\})=[0,1)$ isn't open in ℝ, so I isn't continuous.
    $βˆ€yβˆˆβ„€$, since $I$ is surjective $βˆƒxβˆˆβ„,I(x)=yβ‡’\tilde I([x])=yβ‡’\tilde I$ is surjective
    If $\tilde I([x])=\tilde I([y])$, then $I(x)=I(y)$, so $[x]=[y]$, so $\tilde I$ is injective.
    Suppose $\tilde{I}: ℝ/β„›β†’β„€$ is a homeomorphism when β„€ is endowed with topology 𝒯.
    Since $\tilde I^{-1}$ is continuous, $\tilde I(p(-∞,n])=β„€βˆ©(-∞,n+1)βˆˆπ’―$. So $𝒯=\{βˆ…,β„€\}βˆͺ\{β„€βˆ©(-∞,n+1):nβˆˆβ„€\}$.
    On the other hand, $\tilde I^{-1}(β„€βˆ©(-∞,n+1))=p(-∞,n]$ is open in $ℝ/β„›β‡’\tilde I$ is continuous.
  1. Let $X$ be a topological space and $AβŠ‚X$. On $XΓ—\{0,1\}$ define the partition composed of the pairs $\{(a,0),(a,1)\}$ for $a∈A$, and of the singletons $\{(x,i)\}$ if $x∈Xβˆ–A$ and $i ∈\{0,1\}$. Let β„› be the equivalence relation defined by this partition, let $Y$ be the quotient space $[XΓ—\{0,1\}]/β„›$ and let $p:XΓ—\{0,1\}β†’Y$ be the quotient map.
    1. Prove that there exists a continuous map $f:Yβ†’X$ such that $f∘p(x, i)=x$ for every $x∈X$ and $i∈\{0,1\}$.
    2. Prove that $Y$ is Hausdorff if and only if $X$ is Hausdorff and $A$ is a closed subset of $X$.
  2. Consider the above construction for $X=[0,1]$ and $A$ an arbitrary subset of $[0,1]$.
    Prove that $Y,K=p(XΓ—\{0\})$ and $L=p(XΓ—\{1\})$ are compact, and that $K ∩L$ is homeomorphic to $A$.
    We have thus shown that the intersection of two compact subsets in a space that is not Hausdorff may be non-compact and not closed.
    1. For a class $c$ of the form $c = \{(x,i)\}$, define $f(c) = x$, and for $c$ of the form $c = \{(a,0),(a,1)\}$, define $f(c) = a$
      By definition of product topology, the projection map $f∘p$ is continuous, by Prop 3.9 f is continuous.
    2. If X is Hausdorff and A closed, let $p(x,i)β‰ p(y,j)∈Y$. If $xβ‰ y∈X$, βˆƒopen sets $Uβˆ‹x,Vβˆ‹y$ such that $U∩V=βˆ…$, then $f^{-1}Uβˆ‹(x,i),f^{-1}Vβˆ‹(y,j)$, so $f^{-1}U∩f^{-1}V=f^{-1}(U∩V)=βˆ…$. If $x=y$, then $iβ‰ j,xβ‰ A$, take $(Xβˆ–A)Γ—\{0\},(Xβˆ–A)Γ—\{1\}$. So $Y$ is Hausdorff.
      Conversely, if Y is Hausdorff, for any $x_1β‰ x_2∈X$, since $g:Xβ†’Y,x↦p(x,0)$ is a continuous injection, βˆƒdisjoint open sets $U_1βˆ‹g(x_1),U_2βˆ‹g(x_2)$, so $g^{-1}(U_1)βˆ‹x_1,g^{-1}(U_2)βˆ‹x_2$ are disjoint open sets, and $g^{-1}(U_1)∩g^{-1}(U_2)=g^{-1}(U_1∩U_2)=g^{-1}(βˆ…)=βˆ…$ so X is Hausdorff.
      Suppose $A$ isn't closed. Take $x∈\bar Aβˆ–A$, then $p(x,0)=\{(x,0)\}β‰ p(x,1)=\{(x,1)\}$, βˆƒdisjoint open sets $U_0βˆ‹p(x,0),U_1βˆ‹p(x,1)$, then $f(U_0),f(U_1)$ are open neighborhoods of $x$ in $X$, so $f(U_0)∩f(U_1)$ is an open neighborhoods of $x$, but $x∈\bar A$, so $βˆƒa∈A∩f(U_0)∩f(U_1)$, so $\{(a,0),(a,1)\}∈U_0∩U_1=βˆ…$, contradiction.
  2. $[0,1],\{0,1\}$ are compact, by Thm 2.23 $[0,1]Γ—\{0,1\}$ is compact, since $p$ is continuous, $Y=p([0,1]Γ—\{0,1\})$ is compact, similar for K, L. The only equivalence classes in Y that have both points of the form $(x,0)$ and $(x,1)$ are those with first coordinate in A, so $\{(x,0),(x,1)\}↦x$ is a homeomorphism.
The goal of this exercise is to show there exists an embedding of the real projective plane $ℝP^2$ in $ℝ^4$.
Let $π•Š^2$ denote the unit sphere in $ℝ^3$ given by $π•Š^2=\left\{(x, y, z)βˆˆβ„^3: x^2+y^2+z^2=1\right\}$, and let $f: π•Š^2→ℝ^4$ be defined by $f(x, y, z)=\left(x^2-y^2, x y, y z, z x\right)$. Prove that
  1. $f$ determines a continuous map $\tilde{f}:ℝP^2→ℝ^4$ where $ℝP^2$ is the real projective plane
  2. $\tilde{f}$ is a homeomorphism onto a topological subspace of $ℝ^4$.
  1. Define an equivalence relation on $π•Š^2$: $\left(x_1,y_1,z_1\right)∼\left(x_2,y_2,z_2\right)$ iff $\left(x_2, y_2, z_2\right)=Β±\left(x_1, y_1, z_1\right)$.
    $f$ is continuous. $f(x,y,z)=f(-x,-y,-z)$. By Prop 3.9, $\tilde{f}$ is continuous.
  2. $π•Š^2$ is compact, $ℝ^4$ is Hausdorff, we'll show $f$ is injective. By Prop 3.11, $\tilde{f}$ is a homeomorphism to its image.
    Suppose $(x_1^2-y_1^2, x_1 y_1, y_1 z_1, z_1 x_1)=(x_2^2-y_2^2, x_2 y_2, y_2 z_2, z_2 x_2)$ for $(x_1, y_1, z_1),(x_2, y_2, z_2)βˆˆπ•Š^2$,
    $$\left(x_1^2-y_1^2, x_1 y_1\right)=\left(x_2^2-y_2^2, x_2 y_2\right)β‡’(x_1+y_1i)^2=(x_2+y_2i)^2β‡’(x_1,y_1)=Β±(x_2,y_2)$$Since $x_1^2+y_1^2+z_1^2=x_2^2+y_2^2+z_2^2$ we get $z_1^2=z_2^2$.
    If $(x_1,y_1)β‰ (0,0)$, since $z_1(x_1,y_1)=z_2(x_2,y_2)$, we get $\left(x_2, y_2, z_2\right)=Β±\left(x_1, y_1, z_1\right)$.
    If $(x_1,y_1)=(0,0)$, since $z_1^2=z_2^2=1$, we get $\left(x_2, y_2, z_2\right)=Β±\left(x_1, y_1, z_1\right)$.