Set theory - Chapter 4: Real Numbers

4.1

Let κ\kappa be the cardinal of {f:f is continuous}\{f\in\mathbb{R}^{\mathbb{R}}:f\text{ is continuous}\}. For all kk\in\mathbb{R}, f:xkf:x\rightarrow k is continuous so

κ||=20\kappa\geq|\mathbb{R}|=2^{{\aleph_{0}}}

Let f,gf,g be two continuous functions that have the same restriction on \mathbb{Q}. For all xx\in\mathbb{R}, find a sequence (an)n{(a_{n})}_{{n\in\mathbb{N}}}\in\mathbb{Q}^{\mathbb{N}} such that x=limnanx=\lim_{{n\to\infty}}a_{n}. Then the limit of the sequence f(an)=g(an)f(a_{n})=g(a_{n}) is f(x)=g(x)f(x)=g(x) by continuity. This means that:

κ||=(20)0=20\kappa\leq|\mathbb{R}^{\mathbb{Q}}|={(2^{{\aleph_{0}}})}^{{\aleph_{0}}}=2^{{% \aleph_{0}}}

4.2

Let ζ\zeta be the order-type of \mathbb{R}. For every sequence aa\in\mathbb{N}^{\mathbb{N}} we define the sum of order types τa=a0+ζ+a1+ζ+\tau_{a}=a_{0}+\zeta+a_{1}+\zeta+.... For each xτax\in\tau_{a}, there is some nn\in\mathbb{N} such that xx belongs to the nn-th term. In this nn-th order-type, xx can be viewed as an element of \mathbb{R}. Hence we have a one-to-one map from τa\tau_{a} to ×\mathbb{N}\times\mathbb{R} which is countable by exercise 3.3. Hence τa\tau_{a} is a countable order-type and obviously linear.

Suppose that we have an isomorphism ff between two sets A=A0X0A1X1A2A=A_{0}\sqcup X_{0}\sqcup A_{1}\sqcup X_{1}\sqcup A_{2}... and B=B0Y0B1Y1B2B=B_{0}\sqcup Y_{0}\sqcup B_{1}\sqcup Y_{1}\sqcup B_{2}... of respective order-types τa\tau_{a} and τb\tau_{b}. The elements of AiA_{i} can be written 0i,a,1i,a,(a0-1)i,a0_{{i,a}},1_{{i,a}},...{(a_{0}-1)}_{{i,a}} those of XiX_{i} can be written {pi,a*:p}\{p_{{i,a}}^{*}:p\in\mathbb{R}\} and similarly for Bi,YiB_{i},Y_{i}. Suppose that aba\neq b and let nn be the least index such that anbna_{n}\neq b_{n} or without loss of generality an<bna_{n}<b_{n}.

First note that for any mm\in\mathbb{N} and any pp\in\mathbb{R}, we can not find yBy\in B such that f(pm,a*)<y<f((p+1)m,a*)f(p_{{m,a}}^{*})<y<f({(p+1)}_{{m,a}}^{*}) for otherwise pm,a*<f-1(y)<(p+1)m,a*p_{{m,a}}^{*}<f^{{-1}}(y)<{(p+1)}_{{m,a}}^{*} which is inconsistent with the order-type of XmX_{m}. In particular, for any qq\in\mathbb{R}, qm,a*q_{{m,a}}^{*} is not sent on an element im,bi_{{m^{{\prime}},b}} for otherwise (q-1)m,a*{(q-1)}_{{m,a}}^{*} is necessarily sent to (i-1)m,b{(i-1)}_{{m^{{\prime}},b}}, (q-2)m,a*{(q-2)}_{{m,a}}^{*} to (i-2)m,b{(i-2)}_{{m^{{\prime}},b}}… and so forth until (q-i)m,a{(q-i)}_{{m,a}} which is sent to 0m,b0_{{m^{{\prime}},b}}. Then for p=q-i-1p=q-i-1 we can find a yy leading to a contradiction (we could also have considered p=q+(bm-1-im,b))p=q+(b_{{m^{{\prime}}}}-1-i_{{m^{{\prime}},b}})). This means that f(qm,a*)=qm,b*f(q_{{m,a}}^{*})={q^{{\prime}}}_{{m^{{\prime}},b}}^{*}. From this and using the same method as above we deduce that f(Xm)=Ymf(X_{m})=Y_{{m^{{\prime}}}}.

We shall prove that for all k<nk<n we have f(Ak)=Bkf(A_{k})=B_{k} and f(Xk)=Ykf(X_{k})=Y_{k}. If k<nk<n and these equalities hold for all k<kk^{{\prime}}<k, then necessarily f(0k,a)0k,bf(0_{{k,a}})\geq 0_{{k,b}} and f-1(0k,b)0k,af^{{-1}}(0_{{k,b}})\geq 0_{{k,a}} because ff is one-to-one. But because ff is increasing, the later equality becomes 0k,bf(0k,a)0_{{k,b}}\geq f(0_{{k,a}}) and hence f(0k,a)=0k,bf(0_{{k,a}})=0_{{k,b}}. By induction, we show that f(ik,a)=ik,bf(i_{{k,a}})=i_{{k,b}} for all i<ak=bki<a_{k}=b_{k} and so f(Ak)=Bkf(A_{k})=B_{k} as expected. By the discussion above, f(Xk)=Ylf(X_{k})=Y_{l}. By the induction hypothesis and because ff is one-to-one, lkl\geq k. If l>kl>k then because ff is increasing, ff does not reach any element of YlY_{l}. This contradicts the fact that it is onto. Hence the only possibility is f(Xk)=Ykf(X_{k})=Y_{k}.

As above, f(An)={in,b:i<a0}f(A_{n})=\{i_{{n,b}}:i<a_{0}\} and f(Xn)=Ylf(X_{n})=Y_{l} for some lnl\geq n. Thus since ff is increasing, the element (a0)n,b{(a_{0})}_{{n,b}} (for example) is not reached and this contradicts the fact that is onto. Finally, the map aτaa\rightarrow\tau_{a} is one-to-one and so there are at least ||=20|\mathbb{N}^{\mathbb{N}}|=2^{{\aleph_{0}}} such order-types.

4.3

Let AA be the set of algebraic reals. A\mathbb{Q}\subseteq A so |A|0|A|\geq\aleph_{0}. The set of all integer polynomials can be viewed as a subsets of integer sequences of finite length 1n<ωn\bigcup_{{1\leq n<\omega}}\mathbb{R}^{n}. This set is countable by exercises 3.3 and 3.4, so is well-ordonable. For any xAx\in A, we consider the smallest polynomial PP of which xx is a root. We have PnP\in\mathbb{R}^{n} for some n1n\geq 1 and by the fundamental theorem, PP has roots mnm\leq n roots. a1,a2,,ama_{1},a_{2},...,a_{m}. If kk is such that x=akx=a_{k} then we define f(x)=(P,k)f(x)=(P,k). Hence we have constructed (without the axiom of choice) a one-to-one map from AA to 1n<ωn×\bigcup_{{1\leq n<\omega}}\mathbb{R}^{n}\times\mathbb{N}. Again, by exercises 3.3 and 3.4 this set is countable and hence |A|0|A|\leq\aleph_{0}.

4.4

Let SS be countable. We have sup{|S|,|S|}=|S|+|S|=||=20\sup\{|\mathbb{R}\setminus S|,|S|\}=|\mathbb{R}\setminus S|+|S|=|\mathbb{R}|=2% ^{{\aleph_{0}}}. But because |S|=0<20|S|=\aleph_{0}<2^{{\aleph_{0}}} the only possibility is |S|=20|\mathbb{R}\setminus S|=2^{{\aleph_{0}}}.

4.5

  1. ((i))

    ||=20|\mathbb{R}\setminus\mathbb{Q}|=2^{{\aleph_{0}}} because ||=0|\mathbb{Q}|=\aleph_{0}.

  2. ((ii))

    |A|=20|\mathbb{R}\setminus A|=2^{{\aleph_{0}}} because |A|=0|A|=\aleph_{0} by exercise 4.3.

4.6

Let OO be an open set of \mathbb{R}. We shall define a sequence of open interval Iα=(aα,bα)I_{\alpha}=(a_{\alpha},b_{\alpha}) with rational endpoints and included in OO. The IαI_{\alpha}’s will be pairwise disjoint and we define open sets Oα=β<αIβO_{\alpha}=\bigcup_{{\beta<\alpha}}I_{\beta}. Because every collection of disjoint open nonempty intervals is at most countable (see p. 39) we get a sequence (Iα)α<λ{(I_{\alpha})}_{{\alpha<\lambda}} for some ordinal λ\lambda at most countable.

Hence let α\alpha be an ordinal and suppose that we have contructed IβI_{\beta} for all β<α\beta<\alpha. If OOαO\setminus O_{\alpha} is empty then O=β<αIβO=\bigcup_{{\beta<\alpha}}I_{\beta} and we stop the construction. Otherwise, we start by reducing each IβI_{\beta} and replacing its endpoints by aβ=34aβ+14bβa^{{\prime}}_{\beta}=\frac{3}{4}a_{\beta}+\frac{1}{4}b_{\beta} and bβ=14aβ+34bβb^{{\prime}}_{\beta}=\frac{1}{4}a_{\beta}+\frac{3}{4}b_{\beta} so that the open set U=OOα¯U=O\setminus\overline{O_{\alpha}} is still nonempty. Hence there is a point xUx\in U and a positive number r>0r>0 such that the ball of center xx and radius rr is included in UU. By density of \mathbb{Q} in \mathbb{R}, there are rationals a,ba,b such that x-ra<x<bx+rx-r\leq a<x<b\leq x+r and so a nonempty interval I=(a,b)I=(a,b) with rational endpoints such that IUI\subseteq U. Because there are at most |2|=0|\mathbb{Q}^{2}|=\aleph_{0} such intervals, we can pick one such interval IαI_{\alpha}, without using the axiom of choice.

Finally, reordering the IαI_{\alpha} and adding empty ones (i.e. aα=bαa_{\alpha}=b_{\alpha}) we have shown that every open interval can be written

O=nInO=\bigcup_{{n\in\mathbb{N}}}I_{n}

for InI_{n} an open interval with rational endpoints. Hence there are at most

|(2)|=(02)0=00=20\left|{(\mathbb{Q}^{2})}^{\mathbb{N}}\right|={(\aleph_{0}^{2})}^{{\aleph_{0}}}% =\aleph_{0}^{{\aleph_{0}}}=2^{{\aleph_{0}}}

open sets in \mathbb{R}. The collection of open intervals (0,r)(0,r) for rr\in\mathbb{R} clearly shows that there are exactly 202^{{\aleph_{0}}} many open intervals.

Alternative construction of the decomposition of OO: let (In)n{(I_{n})}_{{n\in\mathbb{N}}} be an enumeration of nonempty open interval with rational endpoints which are included in OO. Let XX the element of nn\in\mathbb{N} for which InI_{n} is minimal i.e. not included in any other ImI_{m}. The (In)n{(I_{n})}_{{n\in\mathbb{N}}} are pairwise disjoint: if InI_{n} and ImI_{m} (n,mXn,m\in X) have nonempty intersection then In,ImInIm=IpIqI_{n},I_{m}\supseteq I_{n}\cap I_{m}=I_{p}\supseteq I_{q} for some pp\in\mathbb{N} and qXq\in X. Hence In=Iq=ImI_{n}=I_{q}=I_{m}. Moreover, we have

O=nXInO=\bigcup_{{n\in X}}I_{n}

4.7

We define for all the closed set (union of 2m<02^{m}<\aleph_{0} closed intervals):

Fm={x=n=1+an3n:an{0,1,2}(n<m,an1)}F_{m}=\left\{x=\sum_{{n=1}}^{{+\infty}}\frac{a_{n}}{3^{n}}:a_{n}\in\{0,1,2\}% \wedge(\forall n<m,a_{n}\neq 1)\right\}

Clearly, the Cantor set is defined by C=mFmC=\bigcap_{{m\in\mathbb{N}}}F_{m} and hence is closed as an intersection of closed sets. Moreover, it is nonempty (for example 0C0\in C). We shall see that it has no isolated point and so is perfect.

Let x=n=1+an3nx=\sum_{{n=1}}^{{+\infty}}\frac{a_{n}}{3^{n}} be a point in CC. If there is n0n_{0} such that an=0a_{n}=0 for all n>n0n>n_{0} then we define for all m,xm=x+23n0+1+mm\in\mathbb{N},x_{m}=x+\frac{2}{3^{{n_{0}+1+m}}}. Otherwise we define xm=n=1man3nx_{m}=\sum_{{n=1}}^{{m}}\frac{a_{n}}{3^{n}}. In both case, xmxx_{m}\neq x and xmCx_{m}\in C for all mm\in\mathbb{N} and limm+xm=x\lim_{{m\rightarrow+\infty}}x_{m}=x, thus xx is not isolated.

4.8

Let PP be a perfect set and P(a,b)=P\cap(a,b)=\emptyset. Define I=[a,b]I_{{\emptyset}}=[a,b]. Then IPI_{{\emptyset}}\cap P is perfect and we can use the proof of theorem 4.5 to construct a one-to-one function from {0,1}ω{\{0,1\}}^{\omega} to P(a,b)P\cap(a,b). Hence |P(a,b)|=𝔠=20|P\cap(a,b)|=\mathfrak{c}=2^{{\aleph_{0}}}.

4.9

Suppose P2P1P_{2}\nsubseteq P_{1} are perfect sets. Let pP2\P1p\in P_{2}\backslash P_{1}. Since P1P_{1} is closed, there is some open interval (a,b)\P1(a,b)\subseteq{\mathbb{R}\backslash P_{1}} such that p(a,b)p\in(a,b). Hence P2(a,b)P_{2}\cap(a,b)\neq\emptyset and by exercise 4.8, |P2(a,b)|=𝔠|P_{2}\cap(a,b)|=\mathfrak{c}. But we have

P2\P1(a,b)\displaystyle{P_{2}\backslash P_{1}}\cap(a,b) =P2\P1(a,b)\displaystyle={P_{2}\cap{\mathbb{R}\backslash P_{1}}}\cap(a,b)
=P2(a,b)\displaystyle=P_{2}\cap(a,b)

Finally 𝔠|P2\P1||P2(a,b)|𝔠\mathfrak{c}\geq|P_{2}\backslash P_{1}|\geq|P_{2}\cap(a,b)|\geq\mathfrak{c}

4.10

For any set PP we have P*P¯P^{*}\subseteq\bar{P} and if PP is closed, P¯=P\bar{P}=P. If moreover PP is perfect and pPp\in P, then a neighborhood of pp contains an open neighborhood (a,b)P(a,b)\cap P of pp which is uncountable by 4.8 and thus pP*p\in P^{*}. Finally, P=P*P=P^{*}.

4.11

For any sets ABA\subseteq B it is easy to see that A*B*A^{*}\subseteq B^{*}. Let PP be a perfect set and FF a closed set, such that PFP\subseteq F. Then P*F*P^{*}\subseteq F^{*} and by 4.10 P*=PP^{*}=P. So PF*P\subseteq F^{*}.

4.12

Let FF be an uncountable closed set and PP the perfect set contructed in theorem 4.6. By exercise 4.11, we have PF*P\subseteq F^{*}. Let aF*a\in F^{*}. For any r>0r>0, we have |B(a,r)F||B(a,r)\cap F| is uncountable by definition of F*F^{*}. But we also have F\PF\backslash P countable so B(a,r)PB(a,r)\cap P is uncountable and a fortiori nonempty. As a consequence, aP¯=Pa\in\bar{P}=P. Finally P=F*P=F^{*}.

4.13

By 4.12, we get F=F*(F\F*)F=F^{*}\cup(F\backslash F^{*}). Let’s see it is the only such decomposition. Suppose F=P1(F\P1)=P2(F\P2)F=P_{1}\cup(F\backslash P_{1})=P_{2}\cup(F\backslash P_{2}) where P1,P2P_{1},P_{2} are perfect. We have F\P1=((F\P2)\P1)(P2\P1)F\backslash P_{1}=((F\backslash P_{2})\backslash P_{1})\cup(P_{2}\backslash P_% {1}). But F\P1F\backslash P_{1} is countable so P2\P1P_{2}\backslash P_{1} is at most countable. By 4.9, this means P2P1P_{2}\subseteq P_{1}. Similarly, we get P1P2P_{1}\subseteq P_{2} and so P1=P2P_{1}=P_{2}.

4.14

Suppose there are open sets UiU_{i} such that

=iUi\mathbb{Q}=\bigcap_{{i\in\mathbb{N}}}U_{i}

In particular, for each ii\in\mathbb{N}, Ui\mathbb{Q}\subseteq U_{i} and because \mathbb{Q} is dense, so is UiU_{i}. Taking the complement, we get

c=iUic\mathbb{Q}^{c}=\bigcup_{{i\in\mathbb{N}}}U_{i}^{c}

Because the UiU_{i}’s are open dense, the UicU_{i}^{c}’s are nowhere dense. Hence the Baire space (which is homeomorphic to c\mathbb{Q}^{c}) is the union of countably many nowhere dense sets. This contradicts the Baire Category Theorem.

4.15

Denote by \mathcal{B} the set of borel sets. Let f:f:\mathbb{R}\rightarrow\mathbb{R} be a continuous function and define SS to be the set of XX\subseteq\mathbb{R} such that f-1(X)f_{{-1}}(X)\in\mathcal{B}. If UU is open, then because ff is continuous f-1(X)f_{{-1}}(X) is open and so belongs to \mathcal{B}. If we have a countable family (Xi)i{(X_{i})}_{{i\in\mathbb{N}}} of elements of SS then we have

f-1(iXi)=if-1(Xi)f_{{-1}}\left({\bigcup_{{i\in\mathbb{N}}}X_{i}}\right)=\bigcup_{{i\in\mathbb{N% }}}f_{{-1}}(X_{i})\in\mathcal{B}

and so iXiS\bigcup_{{i\in\mathbb{N}}}X_{i}\in S. Similarly, if XSX\in S then f-1(Xc)=f-1()f-1(X)cf_{{-1}}(X^{c})=f_{{-1}}(\mathbb{R})\cap{f_{{-1}}(X)}^{c}\in\mathcal{B} and thus XcSX^{c}\in S. Finally, SS is a σ\sigma-algebra which contains the open sets and so SS\supseteq\mathcal{B}. In particular, the preimage of a borel set is a borel set.

4.16

Let f:f:\mathbb{R}\rightarrow\mathbb{R} be a function and SS\subseteq\mathbb{R} the set of points where it is continuous. For all xx\in\mathbb{R} and nn\in\mathbb{N} define the open set

Un=xSB(x,12Nx,n)U_{n}=\bigcup_{{x\in S}}B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}

where for each xSx\in S, Nx,nN_{{x,n}} is the least positive integer such that f(B(x,12Nx,n))B(f(x),12n)f\left(B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}\right)\subseteq B{\left(f(x% ),\frac{1}{2^{n}}\right)}. Such an integer exists because ff is continuous at xx. We shall prove that

S=nUnS=\bigcap_{{n\in\mathbb{N}}}U_{n}

i.e. SS is of type GδG_{{\delta}}. The inclusion from left to right is trivial so let’s consider the other direction. Consider a point ynUny\in\bigcap_{{n\in\mathbb{N}}}U_{n} and a neighborhood VV of f(y)f(y). Let nn be an integer such that B(f(y),12n-1)VB{\left(f(y),\frac{1}{2^{{n-1}}}\right)}\subseteq V. Because yUny\in U_{n}, there exists xSx\in S such that yB(x,12Nx,n)y\in B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}. In particular, f(y)B(f(x),12n)f(y)\in B{\left(f(x),\frac{1}{2^{n}}\right)}. Moreover, if zB(f(x),12n)z\in B{\left(f(x),\frac{1}{2^{n}}\right)}, the triangular inequality gives

|z-f(y)||z-f(x)|+|f(x)-f(y)|<12n+12n=12n-1\displaystyle|z-f(y)|\leq|z-f(x)|+|f(x)-f(y)|<\frac{1}{2^{n}}+\frac{1}{2^{n}}=% \frac{1}{2^{{n-1}}}

Hence if U=B(x,12Nx,n)U=B{\left(x,\frac{1}{2^{{N_{{x,n}}}}}\right)}, we have yUy\in U and f(U)B(f(x),12n)B(f(y),12n-1)Vf(U)\subseteq B{\left(f(x),\frac{1}{2^{n}}\right)}\subseteq B{\left(f(y),\frac% {1}{2^{{n-1}}}\right)}\subseteq V. So ff is continuous at yy i.e. ySy\in S.

4.17

  1. ((i))

    Consider the function f:𝒩×𝒩𝒩f:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N} given by (ai)i<ω×(bi)i<ω𝑓(ci)i<ω{(a_{i})}_{{i<\omega}}\times{(b_{i})}_{{i<\omega}}\overset{f}{\mapsto}{(c_{i})% }_{{i<\omega}} where for all i<ωi<\omega, c2i=aic_{{2i}}=a_{i} and c2i+1=bic_{{2i+1}}=b_{i}. ff is clearly a bijection. We shall prove it is actually a homeomorphism.

    Now pick (a,b)𝒩×𝒩(a,b)\in\mathcal{N}\times\mathcal{N} and define c=f(a,b)c=f(a,b). For any neighborhood VV of cc there exists some nn such that O(c|2n)VO(c_{{|2n}})\subseteq V. Define the open neighborhood U=O(a|n)×O(b|nU=O(a_{{|n}})\times O(b_{{|n}} of (a,b)(a,b). For any (x,y)U(x,y)\in U and 0i<n0\leq i<n, we have f(x,y)2i=xi=ai=c2i{f(x,y)}_{{2i}}=x_{i}=a_{i}=c_{{2i}} and f(x,y)2i+1=yi=bi=c2i+1{f(x,y)}_{{2i+1}}=y_{i}=b_{i}=c_{{2i+1}}. Hence f(x,y)O(c|2n)f(x,y)\in O(c_{{|2n}}). Finally f(U)Vf(U)\subseteq V and ff is continuous.

    Conversely, choose c𝒩c\in\mathcal{N} define (a,b)=f-1(c)(a,b)=f^{{-1}}(c). Let VV be a neighborhood of (a,b)(a,b). There exists some nn such that O(a|n)×O(b|n)VO(a_{{|n}})\times O(b_{{|n}})\subseteq V. Define U=O(c|2n)U=O(c_{{|2n}}). Let zUz\in U and (x,y)=f-1(z)(x,y)=f^{{-1}}(z). For all 0i<n0\leq i<n, we have xi=z2i=c2i=aix_{i}=z_{{2i}}=c_{{2i}}=a_{i} and yi=z2i+1=c2i+1=biy_{i}=z_{{2i+1}}=c_{{2i+1}}=b_{i}. So (x,y)O(a|n)×O(b|n)(x,y)\in O(a_{{|n}})\times O(b_{{|n}}). Finally f-1(U)Vf^{{-1}}(U)\subseteq V and f-1f^{{-1}} is continuous.

  2. ((ii))

    We proceed the same way. Let g:ωω2g:\omega\cong\omega^{2} be a bijection and define the bijection f:𝒩ω𝒩f:\mathcal{N}^{\omega}\rightarrow\mathcal{N} by ((aji)i<ω)j<ω𝑓(ag(k))k<ω{\left({(a_{{ji}})}_{{i<\omega}}\right)}_{{j<\omega}}\overset{f}{\mapsto}{(a_{% {g(k)}})}_{{k<\omega}}. Again, this is a homeomorphism. Let a𝒩ωa\in\mathcal{N}^{\omega} and b=f(a)b=f(a). If VV is a neighborhood of bb then there exists NN such that O(b|N)VO(b_{{|N}})\subseteq V. Now let n=1+max0k<N{π1(g-1(k)),π2(g-1(k))}n=1+\max_{{0\leq k<N}}{\{\pi_{1}{(g^{{-1}}(k))},\pi_{2}{(g^{{-1}}(k))}\}} where π1:𝒩×𝒩𝒩\pi_{1}:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N} and π2:𝒩×𝒩𝒩\pi_{2}:\mathcal{N}\times\mathcal{N}\rightarrow\mathcal{N} are the canonical projection. Define the open neighborhood of aa U=0j<nO((aji)0i<n)×𝒩ωU=\prod_{{{0\leq j<n}}}{O({(a_{{ji}})}_{{0\leq i<n}})}\times\mathcal{N}^{\omega}. Then it’s easy to prove f(U)Vf(U)\subseteq V and so ff is continuous. Conversely, with a=f-1(b)a=f^{{-1}}(b) as above and VV a neighborhood of aa there exists nn such that U=0j<nO((aji)0i<n)×𝒩ωVU=\prod_{{{0\leq j<n}}}{O({(a_{{ji}})}_{{0\leq i<n}})}\times\mathcal{N}^{% \omega}\subseteq V. We define N=1+max0i,j<n{g(i,j)}N=1+\max_{{0\leq i,j<n}}{\{g(i,j)\}} and U=O(b|n)U=O(b_{{|n}}) and open neighborhood of bb. Then f-1(U)Vf^{{-1}}(U)\subseteq V. Thus f-1f^{{-1}} is continuous.

4.18

Let FF be a closed subset of 𝒩\mathcal{N}. We have TF={sω<ω:fF,sf}T_{F}=\{s\in\omega^{{<\omega}}:\exists f\in F,s\subseteq f\}. TFT_{F} is a tree and for sTFs\in T_{F}, there exists fFf\in F and n<ωn<\omega such that s=f|ns=f_{{|n}}. If we define t=s(n,f(n))t=s\cup{(n,f(n))} then tTFt\in T_{F} (because tft\subseteq f) and sts\subsetneq t. So TFT_{F} does not have any maximal node.

Let’s consider the map from the set of closed subsets of 𝒩\mathcal{N} to the set of sequential trees without a maximal node given by FTFF\mapsto T_{F}. By a result of the book, it is obvious that the function is one-to-one: If TF=TGT_{F}=T_{G}, F=[TF]=[TG]=GF=[T_{F}]=[T_{G}]=G. Another (similar) proof: for all ϵ>0\epsilon>0, pick nn such that 12n+1<ϵ\frac{1}{2^{{n+1}}}<\epsilon. We have f|nTF=TGf_{{|n}}\in T_{F}=T_{G}. So there exists some gGg\in G such that f|n=g|nf_{{|n}}=g_{{|n}}. In particular, d(f,g)12n+1<ϵd(f,g)\leq\frac{1}{2^{{n+1}}}<\epsilon. So ff is in G¯=G\bar{G}=G. We do the same the other direction and get F=GF=G.

Let’s prove that the map FTFF\mapsto T_{F} is surjective. We consider TT a sequential tree without maximal node. We define F=[T]={f𝒩,n<ω,f|nT}F=[T]=\{f\in\mathcal{N},\forall n<\omega,f_{{|n}}\in T\}. Let gF¯g\in\bar{F}. For all n<ωn<\omega, O(g|n)FO(g_{{|n}})\cap F\neq\emptyset. So there is fFf\in F such that g|n=f|nTg_{{|n}}=f{|n}\in T. Thus gFg\in F and F¯=F\bar{F}=F i.e. FF is closed. By definition TF={sω<ω:fF,sf}={sω<ω:f𝒩,n<ω,f|nTsf}T_{F}=\{s\in\omega^{{<\omega}}:\exists f\in F,s\subseteq f\}=\{s\in\omega^{{<% \omega}}:\exists f\in\mathcal{N},\forall n<\omega,f_{{|n}}\in T\wedge s% \subseteq f\}. In particular it is clear that TFTT_{F}\subseteq T. Suppose that the previous inequality is strict. Let sTs\in T such that f𝒩,n<ω,f|nTsf\forall f\in\mathcal{N},\forall n<\omega,f_{{|n}}\notin T\vee s\nsubseteq f. Let tt be another element of TT, of length n0n_{0}. Because TT is a tree, we can define for all nn0n\leq n_{0}, tn=t|nTt_{n}=t_{{|n}}\in T. Because TT has no maximal element and is countable, we can construct by induction for all nn0n\geq n_{0}, an element tn+1t_{{n+1}} that extends tnt_{n}. Moreover, because TT is a tree, we can suppose tn+1t_{{n+1}} of length n+1n+1. tnt_{n} is a Cauchy sequence and tnf𝒩t_{n}\rightarrow f\in\mathcal{N}. By construction, f|n=tnTf_{{|n}}=t_{n}\in T. Hence sfs\nsubseteq f and a fortiori sfn0=ts\nsubseteq f_{{n_{0}}}=t. So ss is a maximal element. Contradiction.

4.19

Let PP be a perfect Polish space. Consider a metric such that PP is complete and QPQ\subseteq P a countable dense subset. We denote by B~(x,r)\tilde{B}(x,r) the closed ball of center xx and radius rr. For any sequence s2<ωs\in 2^{{<\omega}}, we shall construct by induction such a ball B~(xs,rs)\tilde{B}(x_{s},r_{s}). We shall take xsQx_{s}\in Q and rsr_{s}\in\mathbb{Q}, so there are at most countably many possibilities and so we won’t need the axiom of choice.

We start by choosing x0Qx_{0}\in Q and setting r0=1r_{0}=1. If BsB_{s} is contructed, then because PP is perfect, the open ball B(xs,rs)B(x_{s},r_{s}) contains at least another point pxsp\neq x_{s}. For ϵ>0\epsilon>0 small enough, the ball B(p,ϵ)B(p,\epsilon) is included in B(xs,rs)B(x_{s},r_{s}) and does not contain xsx_{s}. Then because QQ is dense, there exists some qB(p,ϵ)Qq\in B(p,\epsilon)\cap Q. For a nonzero μ\mu\in\mathbb{Q} small enough, B~(xs,μ)\tilde{B}(x_{s},\mu) and B~(q,μ)\tilde{B}(q,\mu) are two disjoint balls included in BsB_{s} of diameters 2-(length(s)+1)\leq 2^{{-(length(s)+1)}}. Thus we can choose two such balls B~(x(s,0),r(s,0))\tilde{B}(x_{{(s,0)}},r_{{(s,0)}}) and B~(x(s,1),r(s,1))\tilde{B}(x_{{(s,1)}},r_{{(s,1)}}).

Now for each f2ωf\in 2^{{\omega}}, the set nBf|n\bigcap_{{n\in\mathbb{N}}}B_{{f_{{|n}}}} contains exactly one element because PP is complete. We let F(f)F(f) be this element of PP. FF is one-to-one: if f,g2ωf,g\in 2^{{\omega}} are distinct and nn is the least index such that f(n)g(n)f(n)\neq g(n), then F(f)Bf|(n+1)F(f)\in B_{{f_{{|{(n+1)}}}}} and F(g)Bg|(n+1)F(g)\in B_{{g_{{|{(n+1)}}}}} but these two balls are disjoint by construction. We note F-1F^{{-1}} the inverse function defined on the image F(2ω)F(2^{\omega}).

Given f2ωf\in 2^{{\omega}} and ϵ>0\epsilon>0 we take nn such that 2-n<ϵ2^{{-n}}<\epsilon then if gO(f|n)g\in O(f_{{|n}}) i.e. f|n=g|nf_{{|n}}=g_{{|n}}, we have F(f),F(g)B~f|nF(f),F(g)\in\tilde{B}_{{f_{{|n}}}} but this ball is of diameter 2-n<ϵ\leq 2^{{-n}}<\epsilon. Conversely, given F(f)F(2ω)F(f)\in F(2^{{\omega}}) and VV a neighborhood of F(f)F(f), we choose nn large enough so that O(f|n)VO(f_{{|n}})\subseteq V. Then if F(g)F(2ω)F(g)\in F(2^{{\omega}}) is such that F(g)F(g) is in the open neighborhood B(xf|n,rf|n)B(x_{{f_{{|n}}}},r_{{f_{{|n}}}}) of F(f)F(f), we necessarily have g|n=f|ng_{{|n}}=f_{{|n}} and so gVg\in V. Hence both FF and F-1F^{{-1}} are continuous and FF is an homeomophism.

As in exercise 4.7, we consider for each m<0m<\aleph_{0} the set FmF_{m}, union of the of 2m2^{m} closed balls BsB_{s} where ss is of length mm. By construction, F(2ω)=mFmF(2^{\omega})=\bigcap_{{m\in\mathbb{N}}}F_{m} and is closed as an intersection of closed sets. Finally, F(2ω)PF(2^{\omega})\subset P is a closed subset homeomorphic to the Cantor space 2ω2^{\omega}.

4.20

First recall that 2πatan:[0,+)[0,1]\frac{2}{\pi}\operatorname{atan}:[0,+\infty)\rightarrow[0,1] is a homeomorphism and from that we obtain a homeomorphism between [0,+)ω{[0,+\infty)}^{{\omega}} and the Hilbert cube [0,1]ω{[0,1]}^{{\omega}}.

Let PP be a Polish space and {xn,n}\{x_{n},n\in\mathbb{N}\} be a countable dense subset of PP. We define the map f:P[0,+)ωf:P\rightarrow{[0,+\infty)}^{{\omega}} by f(x)=d(x,xn):nf(x)=\left\langle d(x,x_{n}):n\in\mathbb{N}\right\rangle. This map is one-to-one: if xyx\neq y then d(x,y)0d(x,y)\neq 0 and because QQ is dense we can find some xnB(x,d(x,y)2)x_{n}\in B\left(x,\frac{d(x,y)}{2}\right). Then d(x,xn)<d(x,y)2d(x,x_{n})<\frac{d(x,y)}{2} while d(y,xn)|d(y,x)-d(x,xn)|>d(x,y)2d(y,x_{n})\geq\left|d(y,x)-d(x,x_{n})\right|>\frac{d(x,y)}{2} and so f(x)f(y)f(x)\neq f(y). We will note f-1f^{{-1}} its inverse function, defined on its image f(P)f(P).

Let xPx\in P and VV be a neighborhood of f(x)f(x). There are some nn\in\mathbb{N} and ϵ>0\epsilon>0 such that U=0i<nB(xi,ϵ)×[0,+)ωU={\prod_{{0\leq i<n}}{B(x_{i},\epsilon)}}\times{[0,+\infty)}^{{\omega}} is an open subset of VV. If yPy\in P is such that d(x,y)<ϵd(x,y)<\epsilon then 0i<n,|d(x,xi)-d(y,xi)|d(x,y)<ϵ\forall 0\leq i<n,\left|d(x,x_{i})-d(y,x_{i})\right|\leq d(x,y)<\epsilon so f(y)UVf(y)\in U\subseteq V. Conversely, let f(x)f(P)f(x)\in f(P) and ϵ>0\epsilon>0. Because QQ is dense, there is some nn such that xnB(x,ϵ3)x_{n}\in B\left(x,\frac{\epsilon}{3}\right). Then if f(y)f(y) is in the open subset U=[0,+)n-1×B(x,ϵ3)×[0,+)ωU={[0,+\infty)}^{{n-1}}\times B\left(x,\frac{\epsilon}{3}\right)\times{[0,+% \infty)}^{{\omega}} we have in particular d(y,xn)d(y,xn)+ϵ3d(y,x_{n})\leq d(y,x_{n})+\frac{\epsilon}{3} and so d(x,y)d(x,xn)+d(y,xn)2d(x,xn)+ϵ3<ϵd(x,y)\leq d(x,x_{n})+d(y,x_{n})\leq 2d(x,x_{n})+\frac{\epsilon}{3}<\epsilon. Hence both ff and f-1f^{{-1}} are continuous and ff is an homeomophism.

Finally, we have f(P)=nUnf(P)=\bigcap_{{n\in\mathbb{N}}}U_{n} where UnU_{n} are open sets defined by

Un=xP((0i<nB(d(x,xi),2-n))×[0,+)ω)U_{n}=\bigcup_{{x\in P}}\left(\left({\prod_{{0\leq i<n}}{B\left(d(x,x_{i}),2^{% {-n}}\right)}}\right)\times{[0,+\infty)}^{{\omega}}\right)

Hence, any Polish space is homeomorphic to a GδG_{\delta} subspace of the Hilbert Cube.

This page is W3C-compliant - Author: Frédéric WANG
Valid XHTML 1.1 Valid MathML 2.0 Valid SVG Valid CSS