Review of metric spaces. 1. Metric spaces, continuous maps, completeness

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1 (March 14, 2014) Revew of metrc spaces Paul Garrett garrett/ [Ths document s garrett/m/mfms/notes /12a metrc spaces.pdf] We revew bascs concernng metrc spaces from a modern vewpont, and prove the Bare category theorem, for both complete metrc spaces and locally compact Hausdorff [1] spaces. The general dea of metrc space appeared n [Fréchet 1906], and metrc-space structures on vector spaces, especally spaces of functons, was developed by [Fréchet 1928] and [Hausdorff 1931]. See [Deudonné 1981]. Metrc spaces, completeness Completons Bare category theorem 1. Metrc spaces, contnuous maps, completeness [1.1] Abstract noton of metrc space The abstracted dea of a metrc space s of a set wth a noton of dstance on t, d(x, y) = dstance from x to y whch should conform to basc physcal ntuton about dstance. Thus, frst, the only pont y at dstance 0 from a pont x s y = x tself. Second, dstance does not depend on drecton: the dstance from x to y s the same as the dstance from y to x. Thrd, thnkng that the dstance from x to y should correspond to a shortest route of some knd from x to y, the dstance from x to y should be at most the sum of dstances from x to any other ntermedate pont z and then from z to y, gvng the trangle nequalty. That s, a metrc space, d s a set wth a metrc d(, ) : R, a real-valued functon such that, for x, y, z, (Postvty) d(x, y) 0 and d(x, y) = 0 f and only f x = y (Symmetry) d(x, y) = d(y, x) (Trangle nequalty) d(x, z) d(x, y) + d(y, z) A Cauchy sequence n a metrc space s a sequence x 1, x 2,... wth the property that for every ε > 0 there s N suffcently large such that for, N we have d(x, x ) < ε. A pont x s a lmt of that Cauchy sequence f for every ε > 0 there s N suffcently large such that for N we have d(x, x) < ε. A subset of a metrc space s dense n f every pont n s a lmt of a Cauchy sequence n. [1.2] Contnuous maps Generalzng the usual dscusson of contnuty on R n, a map f : from one metrc space to another s contnuous at x f, for every ε > 0, there s δ > 0 such that d (x, x ) < δ mples d (f(x), f(x )) < ε. The map f : s contnuous f t s contnuous at every pont of. [1.2.1] Proposton: A map f : contnuous f and only f, for every Cauchy sequence {x } n convergng to x, lm f(x ) = f ( lm x ) [1] Recall that a topologcal vector space s locally compact f every pont has an open neghborhood wth compact closure. A space s Hausdorff f for any two ponts x, y there are opens U, V such that x U, y V, and U V = φ. 1

2 Paul Garrett: Revew of metrc spaces (March 14, 2014) Proof: On one hand, for contnuous f and x x, gven any ball at x, from some pont onward all the x are nsde that ball. Thus, eventually, f(x ) s close to f(x), so lm f(x ) = f(x). On the other hand, f f were not contnuous at x, then for some ε > 0, for every δ > 0 there s some x nsde the δ-ball at x such that f(x) f(x ) ε. Take the sequence of δ s gven by δ n = 1/n, and let x n be the correspondng bad x. Then x n x but f(x n ) f(x), so f s not contnuous at x. /// [1.2.2] Remark: The property of the proposton s sequental contnuty. It s necessary to assume explctly that {x } has a lmt x, or else make the blanket assumpton that s complete, as ust below. [1.3] Completeness A metrc space s complete f every Cauchy sequence has a lmt. [2] have the advantage that nfntary operatons often make sense. Unsurprsngly, complete metrc spaces The followng standard lemma makes some ntuton explct: [1.3.1] Lemma: Let {x } be a Cauchy sequence n a metrc space, d convergn to x n. Gven ε > 0, let N be suffcently large such d(x, x ) < ε for, N. Then d(x, x) ε for N. Proof: Let δ > 0 and take N also large enough such that d(x, x) < δ. Then for N by the trangle nequalty d(x, x) d(x, x ) + d(x, x) < ε + δ Snce ths holds for every δ > 0 we have the result. /// [1.4] Metrc topologes A metrc space has a natural topology wth bass gven by open balls {y : d(x, y) < r} = (open ball of radus r > 0 centered at x ) That s, a set U s open when around every pont x U there s an open ball of postve radus contaned n U. 2. Completons A not-complete metrc space presents the dffculty that Cauchy sequences may fal to converge. Reasonably, we want to repar ths stuaton, as economcally as possble. [3] Our ntenton s that, for a not-complete metrc space, d, there should be a complete metrc space, d and sometry (dstance-preservng) :, such that every contnuous f : to complete metrc [2] Convergence of Cauchy sequences s more properly sequental completeness. For metrc spaces, sequental completeness mples mples the strongest form of completeness, namely convergence of Cauchy nets, as we wll observe more carefully later. Ths s not mportant at the moment, but wll have mportance for non-metrzable spaces, rarely complete (n the strongest sense), but n practce often at least sequentally complete. A useful form of completeness stronger than sequental completeness but weaker than outrght completeness s local completeness, also called quas-completeness, whch wll play a sgnfcant role later. [3] One tradtonal approach s to mmedately construct a completon as equvalence class of Cauchy sequences n. Ths has the dsadvantage that t does not address the possblty of other constructons yeldng dfferent outcomes. 2

3 Paul Garrett: Revew of metrc spaces (March 14, 2014) space factors through unquely. That s, there are commutatve dagrams [4] of contnuous maps! (for every contnuous???) By chance, ths characterzaton accdentally asks mpossbly much of the completon. The basc problem s that completeness s not a topologcal property, but metrc, as merely-contnuous maps can fal to map Cauchy sequences to Cauchy sequences. [5] That s, the unversal property s plausbly only for some restrcted class of contnuous functons f : preservng Cauchy sequences. An obvous class s non-expandng or contractve f, meanng d (f(x), f(x )) d (x, x ) for all x, x. A somewhat weaker condton on f that promses preservaton of Cauchy sequences s unform contnuty of f :, meanng that, for any ε > 0, there s δ > 0 such that, for all x, x, d (x, x ) < δ mples d (f(x), f(x )) < ε. Unformly contnuous maps preserve Cauchy sequences. [6] Thus, one adequately corrected condton s! (for every contractve ) Such an s a completon of. Thus, we characterze completons by certan of ther nteracton wth other metrc spaces, explanng our expectatons, rather than gvng a constructon wth unclear motvatons. We wll prove that completons exst (by constructng one) and are essentally unque. Before any constructons of completons, we can prove some thngs about the behavor of any possble completon. In partcular, that any two completons are naturally sometrc to each other. Thus, the outcome wll be ndependent of constructon. [2.0.1] Proposton: (Unqueness) Let : and : Z be two completons of a metrc space. Then there s a unque sometrc somorphsm h : Z such that = h. That s, we have commutatve dagram! Z Proof: Frst, take = Z and f : to be the ncluson, n the characterzaton of :. The defnng characterzaton of : shows that there s unque contractve f : fttng nto a [4] A dagram of maps s commutatve when the composte map from one obect to another wthn the dagram does not depend on the route taken wthn the dagram. [5] The nterval [1, + ) and the half-open nterval (0, 1] are homeomorphc by x 1/x, but wth the usual metrc the former s complete whle the latter s not. The Cauchy sequence 1, 2 1, 1 3,... n (0, 1] maps back to the non-cauchy sequence 1, 2, 3,... n [1, ). [6] For unformly contnuous f : and Cauchy {x } n, gven ε > 0, choose δ > 0 so that d (x, x ) < δ mples d (f(x), f(x )) < ε. Choose N large enough so that d (x, x ) < δ for, N. Then d (f(x ), f(x )) < ε for all, N, so {f(x )} s Cauchy n. 3

4 Paul Garrett: Revew of metrc spaces (March 14, 2014) commutatve dagram! f Snce the dentty map s contractve and fts nto ths dagram, the only contractve f fttng nto the dagram s the dentty on. Next, applyng the characterzatons of both : and : Z, we have unque contractve f : Z and g : Z fttng nto! f Z Z! g Then f g : and g f : Z Z ft nto f g Z g f Z By the frst observaton, ths means that f g s the dentty on, and g f s the dentty on Z. Snce the composton of two contractve maps can be an sometry only when both are sometres, f and g are mutual nverses, and and Z are sometrcally somorphc. /// [2.0.2] Remark: A vrtue of the characterzaton of completon s that t does not refer to the nternals of any completon. Next, prove that the mappng-property characterzaton of a completon does not ntroduce more ponts than absolutely necessary: [2.0.3] Proposton: Every pont n a completon of s the lmt of a Cauchy sequence n. That s, s dense n. Proof: Wrte d(, ) for both the metrc on and ts extenson to. Let be the collecton of lmts of Cauchy sequences of ponts n. We clam that tself s complete. Indeed, gven a Cauchy sequence {y } n wth lmt z, let x such that d(x, y ) < 2. It wll suffce to show that {x } s Cauchy wth lmt z. Indeed, gven ε > 0, take N large enough so that d(y, z) < ε/2 for all N, and ncrease N f necessary so that 2 < ε/2. Then, by the trangle nequalty, d(x, z) < ε for all N. Thus, s complete. By the defnng property of, every contractve f : Z to complete Z has a unque extenson to a contractve F : Z fttng nto 4 f F Z

5 Paul Garrett: Revew of metrc spaces (March 14, 2014) Snce s already complete and (), the restrcton of F to gves a dagram f F Z That s, fts the characterzaton of a completon of. By unqueness, s an sometrc somorphsm, so =. /// Now prove exstence of completons by the standard constructon. The ntenton s that every Cauchy sequence has a lmt, so we should (somehow!) adon ponts as needed for these lmts. However, dfferent Cauchy sequences may happen to have the same lmt. Thus, we want an equvalence relaton on Cauchy sequences that says they should have the same lmt, even wthout knowng the lmt exsts or havng somehow constructed or adoned the lmt pont. Defne an equvalence relaton on the set C of Cauchy sequences n, by {x s } {y t } lm s d(x s, y s ) = 0 Attempt to defne a metrc on the set C/ of equvalence classes by d({x s }, {y t }) = lm s d(x s, y s ) We must verfy that ths s well-defned on the quotent C/ and gves a metrc. We have an necton : C/ by x {x, x, x,...} mod [2.0.4] Proposton: : C/ s a completon of. Proof: Grant for the moment that the dstance functon on = C/ s well-defned, and s complete, and show that t has the property of a completon of. To ths end, let f : be a contractve map to a complete metrc space. Gven z, choose a Cauchy sequence x k n wth (x k ) convergng to z, and try to defne F : n the natural way, by F (z) = lm k f(x k ) Snce f s contractve, f(x k ) s Cauchy n, and by completeness of has a lmt, so F (z) exsts, at least f well-defned. For well-defnedness of F (z), for x k and x k two Cauchy sequences whose mages (x k) and (x k ) approach z, snce s contractve eventually x k s close to x k, so f(x k) s eventually close to f(x k ) n, showng F (z) s well-defned. We saw that every element of s a lmt of a Cauchy sequence (xk ) for x k n, and any contnuous respects lmts, so F s the only possble extenson of f to. The obvous argument wll show that F s contnuous. Namely, let z, z, wth Cauchy sequences x t and x t approachng z and z. Gven ε > 0, by contractveness of F, there s N large enough such that d (F ((x r )), F ((x s ))) < ε and d (F ((x r)), F ((x s))) < ε for r, s N. From the lemma above (!), for such r even n the lmt the strct nequaltes are at worst non-strct nequaltes: d (f(x r ), F (z)) ε and d (f(x r), F (z )) ε 5

6 Paul Garrett: Revew of metrc spaces (March 14, 2014) By the trangle nequalty, snce f : s contnuous, we can ncrease r to have d (x r, x r) small enough so that d (f(x r ), f(x r)) < ε, and then d (F (z), F (z )) d (F (z), f(x r )) + d (f(x r ), f(x r)) + d (f(x r), F (z )) ε + ε + ε Snce : s an sometry, d (x r, x r) = d ((x r ), (x r)) d ((x r ), z) + d (z, z ) + d ((x r), z ) so Thus, d (x r, x r) d (z, z ) + 2ε d (F (z), F (z )) d (z, z ) + 4ε (for all ε > 0) Thus, F s contnuous. Grantng that = C/ s complete, etc., t s a completon of. It remans to prove that the apparent metrc on truly s a metrc, and that s complete. Frst, the lmt n attempted defnton d({x s }, {y t }) = lm s d(x s, y s ) does exst: gven ε > 0, take N large enough so that d(x, x ) < ε and d(y, y ) < ε for, N. By the trangle nequalty, d(x, y ) d(x, x N ) + d(x N, y N ) + d(y N, y ) < ε + d(x N, y N ) + ε Smlarly, d(x, y ) d(x, x N ) + d(x N, y N ) d(y N, y ) > ε + d(x N, y N ) ε Thus, unsurprsngly, d(x, y ) d(x N, y N ) < 2ε and the sequence of real numbers d(x, y ) s Cauchy, so convergent. Smlarly, when lm d(x, y ) = 0, then lm d(x, z ) = lm d(y, z ) for any other Cauchy sequence z, so the dstance functon s well-defned on C/. The postvty and symmetry for the alleged metrc on C/ are mmedate. For trangle nequalty, gven x, y, z and ε > 0, let N be large enough so that d(x, x ) < ε, d(y, y ) < ε, and d(z, z ) < ε for, N. As ust above, d({xs }, {y s }) d(x, y ) < 2ε Thus, d({x s }, {y s }) 2ε+d(x N, y N ) 2ε+d(x N, z N )+d(z N, y N ) 2ε+d({x s }, {z s })+2ε+d({z s }, {y s })+2ε Ths holds for all ε > 0, so we have the trangle nequalty. Fnally, perhaps antclmactcally, the completeness. Gven Cauchy sequences c s = {x s } n such that {c s } s Cauchy n C/, for each s we wll choose large-enough (s) such that the dagonal-sh sequence y l = x l,(l) s a Cauchy sequence n to whch {c s } converges. Gven ε > 0, take large enough so that d(c s, c t ) < ε for all s, t. For each, choose () large enough so that d(x, x ) < ε for all, (). Let c = {x,() : = 1, 2,...}. For s, d(c s, c) = lm l d(x sl, x l,(l) ) sup l ( ) d(x sl, x l,(l) ) sup d(x sl, x s,(l) ) + d(x s,(l), x l,(l) ) l 6 2ε

7 Paul Garrett: Revew of metrc spaces (March 14, 2014) Ths holds for all ε > 0, so lm s c s = c, and C/ s complete. /// 3. The Bare category theorem Ths standard result s both ndspensable and mysterous. A set E n a topologcal space s nowhere dense f ts closure Ē contans no non-empty open set. A countable unon of nowhere dense sets s sad to be of frst category, whle every other subset (f any) s of second category. The dea (not at all clear from ths tradtonal termnology) s that frst category sets are small, whle second category sets are large. In ths termnology, the theorem s asserton s equvalent to the asserton that (non-empty) complete metrc spaces and locally compact Hausdorff spaces are of second category. Further, a G δ set s a countable ntersecton of open sets. Concommtantly, an F σ set s a countable unon of closed sets. Agan, the followng theorem can be paraphrased as assertng that, n a complete metrc space, a countable ntersecton of dense G δ s s stll a dense G δ. [3.0.1] Theorem: (Bare category) Let be a set wth metrc d makng a complete metrc space. Or let be a locally compact Hausdorff topologcal space. The ntersecton of a countable collecton U 1, U 2,... of dense open subsets U of s stll dense n. Proof: Let B o be a non-empty open set n, and show that U meets B o. Suppose an open ball B n 1 nductvely chosen. By denseness of U n, there s an open ball B n whose closure B n satsfes B n B n 1 U n For complete metrc spaces, take B n to have radus less than 1/n (or any other sequence of reals gong to 0), and n the locally compact Hausdorff case take B n to have compact closure. Let K = n 1 B n B o n 1 U n For complete metrc spaces, the centers of the nested balls B n form a Cauchy sequence (snce they are nested and the rad go to 0). By completeness, ths Cauchy sequence converges, and the lmt pont les nsde each closure B n, so les n the ntersecton. In partcular, K s non-empty. For locally compact Hausdorff spaces, the ntersecton of a nested famly of non-empty compact sets s non-empty, so K s non-empty, and B o necessarly meets the ntersecton of the U n. /// Bblography [Deudonné 1981] J. Deudonné, Hstory of Functonal Analyss, Mathematcal Studes 49, North Holland, [Fréchet 1906] M. Fréchet, Sur quelques ponts du Calcul fonctonnel, Rend. Crc. mat. Palermo 22 (1906), [Fréchet 1928] M. Fréchet, Les espaces abstrats topologquement affne, Acta Math. 27 (1928), [Hausdorff 1931] F. Hausdorff, Zur Theore der lnearen metrschen Räume, J. de Crelle 167 (1931),