Errata

It is always embarrassing to realize that one’s mind has slipped…

The important bloopers first:

  1. p.120, Definition 5.1.1 (the way-below relation): “for every directed family zi that has a a least upper bound z above y, there is an i in I such that y ≤ zi already.”…  oops! such that x ≤ zi already, not y ≤ zi, as the subsequent explanation (p.121, top) states (found by keith_dr_uk, May 14, 2013).
  2. p.41, Exercise 3.5.8 and p.158, Exercise 5.4.17.  Exercise 5.4.17 is false: the notion of continuous convergence used in Section 5.4 does not specialize to what is called continuous convergence in Exercise 3.5.8.  (Consequence of a remark made by Barth Shiki, July 16, 2013.)
    Call weak continuous convergence the notion of convergence used in Exercise 3.5.8.  Continuous convergence of sequences implies weak continuous convergence.  The converse implication, which is asked for in Exercise 5.4.17, is unknown to me, but seems dubious.
    My preferred fix is to modify Exercise 3.5.8 so as to give the right definition of continuous convergence there: (fn)n ∈ N converges continuously to f if and only if for every point x and every sequence (xn)n ∈ N of points of X that converges to x, for every ε > 0, there are two indices m0, n0 ∈ N such that for all m ≥ m0 and n ≥ n0, d (f(x), fm(xn)) < ε.  Now show that if (fn)n ∈ N converges continuously to f, then f is continuous (as suggested by Barth Shiki), and show (a) through (d), as before.
    One should also modify Exercise 5.4.17, which should now read: Show that the notion of continuous convergence defined in Definition 5.4.15 coincides with the notion of Exercise 3.5.18 for sequences (fn)n ∈ N of continuous maps from a metric space X to a metric space Y.
  3. p.48, Lemma 4.1.10: “A family B of subsets of a topological space X is a base …”: should read “A family B of open subsets of a topological space X is a base …”.  Otherwise, e.g., the family of all subsets of X would be a base… (found, Dec 24, 2013).
  4. p.121, Exercise 5.1.3: “Show that, in N2, x ≪ y iff x ≤ y and x, y ≠ ω” is wrong. You should instead show that the relation ≪ is empty: x ≪ y for no pair of elements x, y. In any case the conclusion that no element is way-below ω remains. (Found by Weng Kin Ho, Sep 04, 2014.)
  5. p.156, Exercise 5.4.12.  You should ignore the last part of the exercise, which asks you to show that Q and the Sorgenfrey line are not consonant.  That is true, but much more involved than the simplistic argument I am proposing.  As far as Q is concerned, see A. Bouziad, Borel measures in consonant spaces, Topology and its Applications 70 (1996):125—138; C. Costantini and S. Watson, On the dissonance of some metrizable spaces, Topology and its Applications 84 (1998):259—268.  As far as the Sorgenfrey line is concerned, see B. Alleche and J. Calbrix, On the coincidence of the upper Kuratowski topology with the cocompact topology, Topology and its Applications 93(3):207—218, 1999.  (found, Jul 10, 2015.)  See also this page for the Sorgenfrey line (added, April 22, 2022.)
  6. p.474, Exercise 9.7.53: the coefficient ring should be C throughout, not Z.  Everything works with Z or C indifferently, except for Hilbert’s Nullstellensatz, which requires the coefficient ring to be an algebraically closed field.  This is the case for C, but certainly not for Z.  In particular, the final sentence should read “Using Kaplansky’s Theorem, show that, up to homeomorphism, the spectrum of Z[X1, …, Xm] is the sobrification of Cm in its Zariski topology.” (found, Nov 17, 2014.)
  7. p.216, Exercise 6.3.10 and p.268, Exercise 7.2.4 and p.285, Exercise 7.3.12: the Sorgenfrey quasi-metric was defined with its arguments swapped.  In other words, what I defined here is the opposite of the Sorgenfrey quasi-metric.  The actual definition is: dl(r,s)=s—r if s≥r, +∞ otherwise.  This is important if you want to retrieve the topology of the Sorgenfrey line in Exercise 6.3.10.
    This has no consequence in the remaining places where the Sorgenfrey quasi-metric is mentioned, except in Exercise 7.3.12: you should replace x by —x, essentially.  That is, the required isomorphism should send (x,r) to (x,—x—r), not to (—x,x—r), you should show that the least upper bound of (xi, ri) is (supi xi, infi ri), and that (x,r) ≪ (y,s) iff x < y and y—x < r—s. (found, Mar 30, 2016.)
  8. p.339, Exercise 7.7.20.  The exercise is wrong.  Indeed, the continuous open images of complete metric spaces are all first-countable (see Exercise 6.3.14), but there are spaces with a continuous model that are not first-countable.  An example is given by {0,1}I for an uncountable set I, which is not-first countable (see Exercise 6.5.9) but is compact Hausdorff, hence has a model by Corollary 8.3.27.  I don’t know of any reasonable condition on Y that would make the result true.  (found by Ng Kok Min, Feb 26, 2017.)
  9. p.387, Lemma 8.4.12.  This only works provided Y is T0.  In other words, the equalizer of two continuous maps from a sober space X to a T0 space Y is sober.  If Y is not required to be T0, then one can get absolutely any subspace A of X as an equalizer (let f map the elements of A to 1 and all others to 2, let g map the elements of A to 1 and all others to 3, where {1,2,3} is given the indiscrete topology).  The proof is unchanged, except that at line 5, ≤ is of course the specialization ordering of Y, not X, and that is an ordering because Y is T0. (found by Zhenchao Lyu and Xiaodong Jia, Feb 08, 2019.)
  10. p.463, Theorem 9.7.12, “the set ↑x ∩ ↑y of common upper bounds of x and y” should read “the set ↓x ∩ ↓y of common lower bounds of x and y“.  The same error occurs at line 5 of the second paragraph of the proof (found by 沈冲 [Shen Chong], May 24, 2019).
  11. p.384, Lemma 8.4.5 and Theorem 8.4.6 are wrong.  The mistake is in line 3 of the proof of Lemma 8.4.5: yes, S preserves coequalizers, but that means that S(q≡) is a coequalizer in the category Sob of sober spaces, not the category Top of topological spaces… and coequalizers in Sob need not be surjective.  In fact every topological space at all can be obtained as a topological quotient of a sober space, and even of a Hausdorff space, by a result of M. Shimrat of 1956.  In general, coequalizers and colimits in Sob are just what I said at the beginning of Section 8.4.1: sobrifications of the corresponding limits taken in Top.  (found by Marcus Tressl, June 19, 2019)
  12. p.85, Exercise 4.7.14.  It is asked to prove that “for every subset A of X, A is the set of all limits of sequences (xi)i ∈ N“.  That should read “for every subset A of X, cl(A) is the set of all limits of sequences (xi)i ∈ N of points of A“.  (found, December 05, 2019)
  13. p.136, Proposition 5.1.60 (Scott’s formula), Corollary 5.1.61, Proposition 5.1.67, Proposition 5.1.69.  It is assumed that Y is a bdcpo for those to apply, but that is not enough.  You either need Y to be a dcpo, or Y to be a bdcpo and B to be cofinal in X (namely, every element of X is below some element of B; that happens if B=X, notably).  The mistake lies in the first lines of the proof of Proposition 5.1.60: “… since {f(b) | b ∈ B, b ≪ x} is directed, using the fact that f is monotonic; and it has an upper bound, namely f(x).  So it has a least upper bound, since Y is a bdcpo.”  If Y is a dcpo, there is no problem: the family  {f(b) | b ∈ B, b ≪ x} is directed, so it has a least upper bound.  If Y is a mere bdcpo, f(x) is meaningless, since f only applies to elements of B.  But, if B is cofinal in X, then x≤b’ for some b’ in B, and then f(b’) is an upper bound of the family, which then has a least upper bound.  (found, February 7th, 2022.)

The less important ones second:

  1. p.25, Proposition 3.25, proof, second paragraph, second line:  “F=⋂i in I Fi” should read “F=⋃i in I Fi” (found by anon1, May 22, 2013).
  2. p.396, Shmuely’s Theorem 8.4.29: I have been told it should be credited to George N. Raney (mentioned by A. Jung, March 22, 2013).  Raney’s celebrated paper (1952) is http://www.jstor.org/stable/2032165, but maybe this is not the right one: I’ve been unable to find it there yet.
  3. p.478, Reference to Nachbin 1965, was reprinted by Robert E. Krieger Publishing Co., not “Robert E. Kreiger Publishing Co.” (May 13, 2013).
  4. p.5, Union Axiom, definition of “∪ x = ∪y ∈ x y” is missing a condition on z: this should be “∪ x = ∪y ∈ x y = {z | ∃ y ⋅ m (y) and z ∈ y and y ∈ x} is a set; i.e., m ({z | ∃ y ⋅ m (y) and z ∈ y and y ∈ x})” (found by Barth Shiki, July 13, 2013).
  5. p.20, Exercise 3.1.5, the hint is missing a power p to the 1-λ denominator: this should be |bi|p / (1-λ)p, not |bi|p / (1-λ) (found by Barth Shiki, July 16, 2013).
  6. p.30, line 14: “d (z→, y→) ≤ 2na/2k“: should be d (z→, y→) ≤ n (2a/2k)p.  Now also correct line 11 so that it say “let k be such that n (2a/2k)p < εp“, and the proof can proceed to the desired conclusion (found by Barth Shiki, July 16, 2013).
  7. p.42, proof of Proposition 3.5.10, line 6: this implicitly uses the inequality d’ (f(x’), f(x)) ≤ ε/3, and this was neither proved nor assumed (found by Barth Shiki, July 16, 2013).  Add the following argument before the sentence starting with “Using the triangular inequality and the axiom of symmetry”: “From the inequality d’ (fn(x), fn(x’)) ≤ ε/3, we easily deduce d’ (f(x), f(x’)) ≤ ε/3, by taking limits over n ≥ nx, with x and x‘ fixed; explicitly, for every ε’ > 0, we have d’ (f(x), fn(x)) < ε’/2 and d’ (f(x’), fn(x’)) < ε’/2 for n ≥ nx large enough, so that the triangular inequality and the axiom of symmetry yield d’ (f(x), f(x’)) ≤ d’ (f(x), fn(x)) + d’ (fn(x), fn(x’)) + d’ (fn(x’), f(x’)) ≤ ε/3 + ε’; since ε’ is arbitrary, d’ (f(x), f(x’)) ≤ ε/3.”
  8. p.43, proof of Theorem 3.5.12, l.10, there is a finite subset Ei such that is Kxi, not K, is included in the union of open balls centered at the points y of Ei (found by Barth Shiki, July 16, 2013).
  9. p.44, l.10, there is an argument missing at the end of this paragraph.  We have shown that fn converged to f uniformly, but we must also show that f is in (calligraphic) K.  This follows from the unused (until now) assumption 3 (found by Barth Shiki, July 16, 2013).
  10. p.61, right before Exercise 4.2.24, remove sentence “If so, then a subset U of X would be Scott-open iff it is upward closed and every chain having a least upper bound in U meets U.”  Not only is it hard to see why it contributes to the argument, but this claim is actually true… see Exercise 4.2.26 (found by Barth Shiki, July 19, 2013).
  11. p.65, remark after Proposition 4.3.9.  More generally, every function from a T1 space to an arbitrary space is monotonic (mentioned by Barth Shiki, July 19, 2013).
  12. p.68, proof of Proposition 4.4.7, l.3, “such that Ui1, …, Uin ⊆ U” should be “such that Ui1, …, Uin ⊆ Ui” (found by Barth Shiki, July 19, 2013).
  13. p.71, proof of Proposition 4.4.17, lines 10-11, replace “F’ is compact, so extract a finite subcover (Vx)x ∈ E of F’” by “F is compact, so extract a finite subcover (Ux)x ∈ E of F” (found by Barth Shiki, July 19, 2013).
  14. p.312, Fact 7.5.23, there are extraneous ‘Y‘ subscripts to the categories YCQMet and YCQMetu; they should be ignored (found, April 9, 2016).
  15. p.34, before Theorem 3.4.5, and p.209, Definition 6.2.2: c-Lipschitz maps are defined in the expected way, as those maps f such that d(f(x),f(y)) ≤ c.d(x,y).  This is however meaningless when c=0 and d(x,y)=+∞.  This is repaired by taking the convention that 0.+∞=+∞.  That convention is important: with that convention, f is c-Lipschitz if and only if Bc(f) is monotonic, and f is c-Lipschitz Yoneda-continuous if and only if Bc(f) is Scott-continuous (Proposition 7.4.38), but the equivalence would fail with any other convention (found, Sept. 14, 2016).
  16. p.293, Lemma 7.4.17.  The proof needs to be fixed (the Lemma holds).  The problem stems from the fact that the family (ri)i ∈ I is implicitly assumed to be filtered, which it need not be.  This is repaired by noticing that the family (ri+si)i ∈ I is filtered.  Here is the amended proof. Let ((xi, ri), si)i ∈ I, ⊑ be a Cauchy-weighted net in B(X, d).  If i ⊑ j then d+ ((xi, ri), (xj, rj)) ≤ si–sj, and this implies both si≥sj, and d (xi,xj) ≤ ri+si–(rj+sj); in particular, ri+si≥rj+sj.  It follows that (si)i ∈ I and (ri+si)i ∈ I are filtered families. Let r=infi ∈ I (ri+si). Then (xi, ri+si–r)i ∈ I, ⊑ is Cauchy-weighted in X, d.  Let x be the d-limit of (xi)i ∈ I, ⊑.  By Lemma 7.4.9, d(x, y)=supi ∈ I (d(xi, y)–ri–si+r)), a directed supremum, for every y in X.  By a similar argument as in the book, it follows that for every formal ball (y, s), d+ ((x, r), (y, s)) = supi ∈ I (d+ ((xi, r), (y, s)) – si), showing that (x, r) is a d+-limit of (xi, ri)i ∈ I, ⊑.  (found, Dec 1, 2016.)
  17. p.267, Proposition 7.1.20: “Every limit of any net (xi)i ∈ I, ⊑ is a dop-limit” should read “Every limit of any net (xi)i ∈ I, ⊑ in X, dsym is a dop-limit”, as the proof makes clear.  (found, Jan 17, 2017.)
  18. Section 7.5, p.311 and subsequent pages, the formal ball completion.  One should give credit to Steve Vickers for that construction, see the reference below.  I could say that I did not know of that when I wrote the book, but that would not be true: Achim Jung had very kindly directed me to Steve’s work after I gave a talk on the subject at the Topology, Algebra, Categories and Logic conference in Marseilles in the summer of 2010… and I forgot about it.  (found back, July 2016; inserted here, Jan 17, 2017.)
  19. p.198, Proposition 5.7.12.  The assumption that X is core-coherent is not needed here.  The result holds for every topological space X at all, provided that Y is a continuous poset.  As a result, Proposition 5.7.13 and Proposition 5.7.14 also hold without any need for core-coherence.
    The use of core-coherence can be circumvented as follows.  We first show that, if U is open and U ⋐ f-1(↟y) then U ↘ y ≪ f.  The proof is as in the book, only simpler, since we do not need to handle several elementary step functions.  Then we recall that a step function is a supremum of finitely many elementary step functions, and we note that, in any poset, if finitely many elements fi are way-below a given element f, and they have a supremum, then that supremum is also way-below f; the proof of the latter is an easy exercise.  (found, Jan 18, 2017.)
  20. p.432, comment between Proposition 9.5.28 and Proposition 9.5.29: general limits of spectral spaces in Top are not spectral in general.  This is true, but the proposed argument is wrong. Since all spectral spaces are sober, and limits of sober spaces in Top, a limit of spectral spaces must be sober.  In particular, we don’t obtain all T0 spaces as such limits (found by Paweł Bilski, May 23, 2017).  To support the claim that limits of spectral spaces in Top need not be spectral, instead, consider some (non-Noetherian) spectral space X, and some open subset U of X that is not compact.  Then the equalizer of χU and of the constant 1 map (both from X to S) is U itself, which cannot be spectral since it is not even compact.  For an example of such a situation, take X=P(N) with the Scott topology of inclusion, and U be the set of non-empty subsets of N, i.e., the union of the open subsets ↑{n}, n in N.
  21. p.440—441, several places where one should assume the spaces to be T0.  At the bottom of p.440, “In this case, each of s, r determines the other one” is true only if both X and Y are T0.  On p.441, Lemma 9.6.5, the “unique associated section” is not unique unless Aj is T0, so one should assume given a projective system of T0 spaces, not of general topological spaces (found, Feb 15, 2018).
  22. p.464, proof of Theorem 9.7.12, the paragraph starting with “We claim that each En contains exactly one element” contains a bug.  Precisely, the statement “So the sequence ↓(↓x ∩ En) ⊇ ↓(↓x ∩ En+1) ⊇ · · · can only contain finitely many distinct subsets” is faulty, since that sequence need not be descending in general.   One must replace ↓(↓x ∩ En) by ↓x ∩ ↓En, and check that this is the downward closure of a finite set—a direct consequence of property W (found by 沈冲 [Shen Chong], May 24, 2019).
    Explicitly, one should replace the faulty paragraph by the following one:

    • We claim that each En contains exactly one element. Else, let n be minimal such that En contains at least two elements. For each x ∈ En, {x} has strictly less elements than En, hence is good. Noting that for every m ≥ n, ↓x ∩ ↓Em can be written as ↓Fxm for some finite set Fxm, using property W, and noting that we can take Fxn = {x}, the sequence ↓x =  ↓Fxn ⊇ ↓x ∩ ↓En+1 = ↓Fx(n+1) ⊇ … ⊇ ↓x ∩ ↓Em = ↓Fxm ⊇ … can only contain finitely many distinct subsets. It follows that there is an Nx ≥ n such that ↓x ∩ ↓Em = ↓x ∩ ↓ENx for every m≥Nx.  Since En is finite, there is a natural number N above every Nx, x ∈ En.  For every m≥N, then, ↓x ∩ ↓Em = ↓x ∩ ↓EN, both sides being equal to ↓x ∩ ↓ENx.  We then have ↓Em = ↓En ∩ ↓Em (since ↓En ⊇ ↓Em) = ∪x ∈ En (↓x ∩ ↓Em) = ∪x ∈ En (↓x ∩ ↓EN) = ↓En ∩ ↓EN = ↓EN.  Taking m=N+1, this contradicts ↓EN ⊃ ↓EN+1.
  23. p.465, line 4 (inside Exercise 9.7.29), the closed subset (curly) F of X* should be assumed non-empty, otherwise the complement does not have the right form (found by 沈冲 [Shen Chong], May 31, 2019).
  24. p.186, before Fact 5.6.14, “Since C is a left adjoint, it preserves all colimits, whence”… but C is a right adjoint, not a left adjoint.  Fact 5.6.14 nonetheless holds (found by Xiaodong Jia, July 09, 2019), but the sentence we just cited must be replaced by the following more complex justification:
    • Since the inclusion functor from TopC into Top is a left adjoint, it preserves all colimits, so all colimits in the former must be computed as in Top. The latter are quotients of coproducts taken in Top (Example 4.12.12). Let us check that the result is C-generated. For every family (Xi)i ∈ I of C-generated spaces, we check that X = ∐i ∈ I Xi is C-generated: given any subset U of X such that k-1(U) is open in C for every C-probe k : C → X, in particular k-1(ιi-1(U)) is open in C for every C-probe k : C → Xi and every i ∈ I, since then ιi o k is a C-probe, too; since Xi is C-generated, ιi-1(U) is open in Xi for every i ∈ I, and that is enough to conclude that U is open in X. Similarly, if X is C-generated and q : X → Y is quotient, then Y is C-generated: given any subset V of Y such that k-1(V) is open in C for every C-probe k : C → Y , then this holds in particular for all the C-probes q o k with k : C → X; so, using the fact that X is C-generated, q-1(V) is open in X, and because q is quotient, V is open in Y. It follows that all colimits of C-generated spaces in Top are C-generated, and are the corresponding colimits in TopC.
  25. p.343, Exercise 8.1.4, “Show that  […] the complete Heyting algebras are exactly the Cartesian-closed, small, thin categories.”  That is wrong.  A complete Heyting algebra is, in particular, a poset, not just a preordered set.  As a category, it must therefore be separated, in the sense that any two isomorphic objects are equal.  The statement should therefore read “Show that  […] the complete Heyting algebras are exactly the Cartesian-closed, small, separated thin categories.” (found, March 06, 2021.)
  26. p.471, proof of Lemma 9.7.45: “Take any element t of Vnqiq = πnqiq \(π0 ∪ π1 ∪ ··· ∪ πn0–1 ∪ πn0  ∪ ··· ∪ πnq–1), in particular t is in πnqiq hence in π0 ∪ π1 ∪ ··· ∪ πn0–1 ∪ πn0i0 ∪ πn1i1 ∪ ··· ∪ πnq–1iq–1. Since t is not in π0 ∪ π1 ∪ ··· ∪ πn0–1, t must be in some πnrir \(π0 ∪ π1 ∪ ··· ∪ πn0–1 ∪ πn0 ∪ ··· ∪ πnq–1) with r <q, hence in Vnrir.”  The very last claim is only true if nr<nq, in which case t is indeed in Vnrir=πnrir \(π0 ∪ π1 ∪ ··· ∪ πn0–1 ∪ πn0 ∪ ··· ∪ πnr–1), but we don’t know that.  (found by Aliaume Lopez, November 02, 2022.)  This can be repaired by requiring the indices nr to be strictly increasing in r.  Just replace the first paragraph of the proof (bottom of p.470) by:
    • Assume Y’ were not Noetherian. By Lemma 9.7.15, there would be a bad sequence Vn0i0, Vn1i1, …, Vnqiq, … Note that all the pairs (nq,iq) are pairwise distinct: if we had (nr,ir)=(nq,iq) with r<q, then we would have Vnrir=Vnqiq, contradicting the fact that the sequence is bad. We build a sequence of indices kj by induction on j ∈ N such that kj<kj+1 and nkj<nkj+1 as follows. First, we pick k0 arbitrarily. Assuming that kj has been built, we note that there are only finitely many pairs (nq,iq) such that nq≤nkj, since they are all pairwise distinct; namely, at most ∑n=0kj mn. Among the infinitely many pairs (nq,iq) such that q>kj, there must therefore be at infinitely many such that nq>nkj. We pick one and call kj+1 the index q we have just found. This being done, we obtain a subsequence Vnkjikj (j ∈ N) of our original sequence. Any subsequence of a bad sequence is bad, and this one additionally satisfies nkj<nkj+1 for every j ∈ N. Replacing the original sequence Vn0i0, Vn1i1, …, Vnqiq, … by that subsequence if necessary, we may therefore safely assume that nr<nq for all r<q.

Things that I once thought were mistakes but are not:

  1. p.311, Definition 7.4.72.  I proposed to define a Yoneda-complete quasi-metric space as being continuous if and only if its dcpo B(X, d) of formal balls is continuous.  For conformance with Kostanek and Waszkiewicz’s paper cited there, I realized that one should really define a Yoneda-complete quasi-metric space X, d as being continuous if and only if its dcpo B(X, d) of formal balls is continuous and its way-below relation ≪ is standard, in the sense that for every non-negative real number a, (x, r) ≪ (y, s) if and only if (x, r+a) ≪ (y, s+a).  The standardness condition I just stated is a simplification of their characterization, but is equivalent.  Note that (x, r) ≪ (y, s) always implies (x, r+a) ≪ (y, s+a).   (found, April 23, 2016.)
    It turns out that the converse implication also holds, so that was no mistake after all (proved by Ng Kok Min, August 24, 2016.)

Typos:

  1. p.10, Section 2.3.3, first line: “A map f from a poset X to a quasi-ordered set Y is monotonic iff, for every x, x’ ∈ X with x ≤ x’, f (x) ≤ f (x’)”; there is no need to assume X to be a poset here, and the definition should instead start with “A map f from a quasi-ordered set X to a quasi-ordered set Y“.  All the more so as the next sentence starts with “When X is a poset”… (found by Barth Shiki, July 13, 2013).
  2. p.21, line -3: “and d(y,yn) < ε for n large enough” should read “and d(y,xn) < ε for n large enough” (found by Barth Shiki, July 16, 2013).
  3. p.22, last line, “metric point” should read “metric space” (found by anon1, May 22, 2013).
  4. p.26, l.7, sentence between parentheses: the formula defining xn has four occurrences of x; these should be n, not x (ln (n+2), cos (n+2), sin (1/2 n+3/2)) (found by Barth Shiki, July 16, 2013).
  5. p.33, line 5, “for every n ≥ ni, di (xi, xmi) < ε” should read “for every n ≥ ni, di (xi, xmni) < ε”; and line 6, “di (x→, x→m) < ε” should read “di (x→, x→mn) < ε” (found by Barth Shiki, July 16, 2013).
  6. p.33, line -6: “every xn with n ≠ n1” should read “every xn with n ≥ n1” (found by Barth Shiki, July 16, 2013).
  7. p.34, line 7: “Then xn0– — xn0+ ≤ ε/2″ should be “Then xn0+ — xn0– ≤ ε/2″ (found by Barth Shiki, July 16, 2013).
  8. p.36, last four lines, all the occurrences of xn should read xnn, all occurrences of xm should read xmm (4 occurrences of each; found by Barth Shiki, July 16, 2013).
  9. p.45, l.1, “So Corollary 3.5.13 applies” should read “So Theorem 3.5.12 applies” (found by Barth Shiki, July 16, 2013).
  10. p.64, proof of Proposition 4.3.6, l.4, “(xn)n ∈ N” should read “(fn (⊥))n ∈ N” (found by Barth Shiki, July 19, 2013).
  11. p.120, Definition 5.1.1 (the way-below relation): duplicate “a” (May 14, 2013).
  12. p.177, last paragraph, in the definition of an adjoint functor, the functor “f” should be “F” (found, Feb 21, 2015).
  13. p.306, l.2 (Exercise 7.4.54): remove “and ηEHB(X,d)“, which is a spurious duplicate of the subsequent “and the section ηEHB(X,d)” (August 21, 2013).
  14. p.479, reference to “Weihrauch, K. and Schneider, U. 1981” should read “Weihrauch, K. and Schreiber, U. 1981” (August 26, 2015).
  15. p.250, rightmost diagram at the top: the leftmost vertical arrow should be labeled cC,A ⊗ idB, not cC,A ⊗ idC (February 21, 2018).
  16. p.462, line 9: “every open rectangle is of the form (X1 – ↓E1) × (X2 × ↓E2) with E1 and E2 finite” should read “every open rectangle is of the form (X1 – ↓E1) × (X2 – ↓E2) with E1 and E2 finite” (found by 沈冲 [Shen Chong], May 31, 2019).
  17. p.423, line 2: “for each x ∈ X such that g(x)>ε”: this should be ≥, not > (found, April 06, 2020).
  18. p.364 (Exercise 8.2.48): “This result is due to Lawson (1997a), where the rounded topology was called the quasiScott topology”: replace “quasiScott” by “pseudoScott” (found, May 16, 2020).

Clarifications:

  1. p.100, Example 4.8.2, Baire space.  It is N that should be taken with the discrete topology.  Baire space is the product of countably many copies of the latter.  Its topology is far from discrete, although it was built up as a product of discrete spaces (July 16, 2013).
  2. Discussion, p.21.  I am not stating which metric I’m using here.  I meant the L2 metric of Exercise 3.1.3.  However, we shall see that as far as convergence is concerned, the L1 metric, the L2 metric, the Lp metric (with p ≥ 1), the L∞ metric (a.k.a., the sup metric) will all yield similar results: they all define the same topology, and notions of convergence only depend on the topology (mentioned by Barth Shiki, July 16, 2013).
  3. p.45, Exercise 3.5.15.  I am not stating which metric I’m using on Rm.  This can be any of the Lp metrics (with p ≥ 1), as required to apply the Borel-Lebesgue Theorem (Proposition 3.3.4).  As in the previous point, we shall realize later that this does not matter (mentioned by Barth Shiki, July 16, 2013).
  4. p.65, Exercise 4.3.7.  The ordering on R x R, or on R+ x R+, is the product ordering.  That can be inferred from the fact that this is the only one we have defined on products, but is ambiguous anyway (mentioned by Barth Shiki, July 19, 2013).
  5. p.177, bottom.  I have written “The definition above makes it clear that an equivalence is just an adjunction F ⊣ G whose unit and counit are iso.”  This is wrong: an adjunction whose unit and counit are iso is called an adjoint equivalence in the literature.  The discussion that I referred to shows that every adjoint equivalence is an equivalence, but an equivalence need not be an adjoint equivalence.  However,  the two concepts are very close.  Given an equivalence between F and G, we can turn it into an adjoint equivalence by changing the natural isomorphism between FG (and leaving the other one unchaged), or conversely; each can be done in a unique way, in fact. (found, Feb 21, 2015).
  6. p.368, proof of Theorem 8.3.10, that every core-compact sober space is locally compact.  At line 3, I am saying “By interpolation in O(X), since this is a continuous dcpo, there is an open subset U1 such that Uω ⋐ U1 ⋐ U0“. Interpolation refers to Proposition 5.1.15, and I should have said so explicitly.
  7. p.137, Corollary 5.1.61 (of Scott’s formula).  In the proof, one finds the equality “supb ∈ B, b ≪ b’ f(b) = f(supb ∈ B, b ≪ b’ b)”: the supremum on the right, supb ∈ B, b ≪ b’ b, is computed in X, but we have assumed that f is Scott-continuous from B to Y, meaning that f preserves directed suprema as computed in B, not X; fortunately, any directed supremum of elements of B, as computed in X, that happens to be in B, is also a supremum as computed in B, but beware that suprema computed in B may fail to be suprema of elements of B as computed in X. (found, September 19th, 2023.)

Simplifications and Improvements

  1. The proof of Lemma 8.3.42 is overly complicated, and one can at the same time prove something more general: every space Y (not necessarily sober) such that O(Y) is prime-continuous is a c-space.  Therefore the c-spaces are exactly the spaces whose lattice of open sets if prime-continuous, or equivalently, completely distributive (see Exercise 8.3.16).  See Appendix B of this post for an elementary proof (found, January 22nd, 2022.)

— Jean Goubault-Larrecqjgl-2011