Q     We saw in class how to prove the "quadrilateral inequality" from the triangle inequality, and indicated how to inductively obtain pentagon, hexagon, etc. inequalities as well. Can we conversely deduce the triangle inequality from the quadrilateral inequality?

A     Yes: if we know that

d(x,y) <= d(x,r) + d(r,s) + r(s,y)
for all x,r,s,y in X, then by setting s=y and using the axiom d(z,z)=0 we deduce
d(x,y) <= d(x,r) + d(r,y) + d(y,y) = d(x,r) + d(r,y).
(We could also have taken r=s or r=x instead of s=y.) Thus the quadrilateral and triangle inequalities are equivalent, assuming the axiom d(z,z)=0. In the same way we can show that the triangle inequality is equivalent to the pentagon inequality, and more generally to the n-gon inequality for each n>3.
Q     What are "NB" and "cf." on the first problem set? These notations were never defined.

A     These are not mathematical notations. They're abbreviated Latin phrases that are occasionally used in English text, analogous to "etc." ("et cetera": and so forth), "i.e." ("id est": that is), and "e.g." ("exempli gratia": for example), but possibly less familiar. Other such abbreviations that you may see in Math 55 and (more likely) elsewhere are: "et al.", "loc. cit.", "op. cit.", "q.v.", and "viz.". Look them up in your English dictionary when they arise.

Q     What was that homework grading policy again?

A     Most (about 2/3) of your grade will be based on your performance on the problem sets, each of which counts equally. On each problem set that you hand in late without a good excuse, you will be penalized by having your score for that problem set multiplied by the factor (2/3)N, where N is the number of late days (rounded upwards to an integer). After this adjustment, your worst problem grade of the semester will be dropped before computing your total homework score. You have one free (unexcused, un-penalized) late homework day for the semester. Use it wisely. :-)

Q     I didn't quite catch the definition of absolut [sic] convergence.

A     Footnote 2 of the '95-'96 Math 55 T-shirt appears on page 2 of this PDF file of the back of the shirt. The fake final exam itself is on page 1.

Q     More seriously: What's an isometry? I need this for Problem 4 of the first problem set. The index to Rudin says ``Isometry, 82, 170'', but I cannot find a definition on either page.

A     An isometry is a distance-preserving bijection between metric spaces. This notion is defined on page 2 of the first topology handout, on the third and fourth lines after the displayed equation in the middle of the page.

Warning: Rudin does give a brief definition on page 82 (towards the end of Exercise 24) and again on page 170 (in the middle of Exercise 24 of that section), but that definition is either incorrect or outdated, because it does not require that the map be a bijection. A distance-preserving mapping -- that is, a function f between metric spaces satisfying d(f(x), f(y)) = d(xy) for all xy -- would nowadays be said to be ``an isometry to its image''. Note that such an f is automatically an injection (why?).

Q     I think I found an error in the handout / problem set / website / textbook.

A     You may well be right. I'm not perfect, and even Rudin errs on occasion (as already noted). Please let me know, and I'll correct the error in class and/or by updating the webpage --- assuming it really is an error. But first, please check the class website: the error may have already been spotted and corrected.

Q     How am I supposed to remember that a lower-case Greek zeta has two strokes and xi has three, and not the other way around?

A     In general, a lower-case letter is a stylized version of what you get when you try to draw an upper-case (capital) letter quickly. The Greek capital Zeta looks just like the Roman capital letter Z that we all know; if you try to draw it quickly you'll get something like the two-stroke lower-case zeta. The Greek capital Xi is not an X shape (that's Chi, as in ``chiasmus'' and the Cyrillic kh letter); it consists of three horizontal line segments, the middle one shorter than the other two. Draw that shape quickly from top to bottom, and you'll get the three-stroke lower-case xi.

Or, you may remember that 3 goes with xi rather than zeta by using a recent blockbuster movie as a mnemonic: it's XXX, not ZZZ.

[If your browser has the Symbol font installed, then X,Z will be capital Xi and Zeta, and x,z will be the lower-case versions.]

Q     Where did that integral formula for n! come from?

A     It's due to Euler (who is also the source of the ``integral formula'' for 0!-1!+2!-3!+4!-+... that evaluates to 0.596...). See Rudin, page 192, Theorem 8.18. The formula can be proved inductively by repeated integration by parts. Inna Z. notes that it can also be obtained starting from the formula 1/a = int(e-ax dxx=0..oo) by differentiating n times under the integral sign and then setting a=1 -- though this approach requires more justification, especially here where we're differentiating an improper integral under the integral sign.

Q     We showed with little difficulty that if E is a subset of a metric space X then E is closed if and only if it satisfies the following criterion:
Let {pn} be any sequence of elements of E. If there exists p in X such that pn->p then p is in E.
Is the same thing true in an arbitrary topological space?

A     Good question. The answer is no: every closed set satisfies this criterion, but some topological spaces X have non-closed subsets E that satisfy it as well. The proof of the first statement is easy; constructing a counterexample to the converse assertion takes some more work.

Suppose first that E is closed, and let {pn} be a sequence of elements of E that converges to some p in X. If p were not in E then (since E is closed) there would be an open set U containing p but disjoint from E. Since p is a limit of {pn}, U must contain pn for all large enough n. But U is disjoint from E, and therefore can contain no pn at all. This contradiction proves that p is in E as desired.

If the converse were true, we would be able to start from any non-closed set E and find a point p outside E and a sequence {pn} in E such that every open U containing p also contains all but finitely many pn. For a metric space we did this using in effect the fact that the sequence of open sets Un=B1/n(p) has the the following property: every open U that contains p must contain some Un. Such a sequence does not exist in a general topological space, so we may well expect counterexamples. I can construct a counterexample, but only using some notions that we have not yet developed yet, and might have no occasion to develop soon...

Q'     Can you please show me that counterexample?

A     If you insist... I'll outline one counterexample without attempting to give all the details.

Let X be the space {0,1}R of functions from R to {0,1}. [The only fact we'll need about the set R is that it is uncountable.] We equip X with the ``product topology'' relative to the discrete topology on the 2-element set {0,1}. This is the smallest topology that makes all the projection maps from X to {0,1} continuous. An even more direct statement (but one that still takes work to fully unwind) is that X carries the smallest topology in which {f | f(z)=0} and {f | f(z)=1} are open for every real z. You can now easily check that X is Hausdorff. It turns out that a sequence {fn} in X converges to f in this topology if and only if it convergences to f pointwise; that is, if and only if for each real z there exists N such that fn(z)=f(z) for all n>N. Now let E be the set of functions such that f-1(1) is countable. [NB In our terminology a finite set is considered countable.] It's easy to check that E satisfies our criterion; the key fact is that a union of countably many countable sets is itself countable. But it turns out that E is not closed in X.

Q''     So can anything be salvaged of the converse statement for topological spaces that need not be metric?

A     One approach is to generalize the notion of a sequence {pn}, replacing the index set {1,2,3,...} by an arbitrary poset [partially ordered set] satisfying the condition that for all n,n' in the poset there exists N such that N>n and N>n'. One can then fix p and use, instead of the sequence of open sets Un=B1/n(p), the generalized sequence of all open sets containing p, with "U>V" meaning that U is a subset of V. We can then prove the closure criterion as we did in a metric space.

We'll probably not pursue generalized sequences further in Math 55.

Q     To get back to earth: What was that example mentioned in class of a sequence in some topological space with a non-unique limit point?

A     Let X be an arbitrary set endowed with the "indiscrete topology" in which the only open sets are the empty set and X itself. Then any sequence converges, and has every point of X as a limit! In particular, if X is not a zero- or one-point set then every sequence has non-unique limit points.

Q     What does ``extension'' mean in the last problem of the second set?

A     Let X,Y be any sets, and let E be any subset of X. Suppose  f  is a function from E to Y. An extension of  f  to (a function on) X is any function g from X to Y such that g(x)=f(x) for all x in X. In other words, it is a function from X to Y whose restriction to E recovers f.

If you want to get fancy, consider ``restriction from X to E'' as a map from YX={functions from X to Y} to YE={functions from E to Y}; then an extension of  f  is a preimage of  f  under this map.

Q     I can hardly read the PS and PDF files on my computer screen, even when I magnify them to almost twice [that is, 22 times] their size.
Q'     I can't read the PDF files at all: they turn up as blank pages!

A     Try the new PDF files, generated directly by pdflatex instead of the composite function ps2pdf(dvips(latex(·))), and accessible by links marked PDF' rather than PDF. Preliminary reports indicate that these files solve both problems.

The PDF's still can't be as clear as printed handouts because computer screens have lower resolution than printers. But each of the three versions (PS, PDF, or PDF') does contain all the information needed to recover a good copy of the handout or problem set by sending it to a printer.

Q     I didn't catch the distinction between the direct sum and product of a (possibly infinite) family of vector spaces Vi. I can't look it up in Axler, since Axler considers only finite sums of vector spaces. Which is which?

A     The product (\prod_i  V_i  in TeX) is the usual Cartesian product, and consists of all assignments {vi} of a vector vi in Vi for each index i. The vector-space operations of zero, additive inverse, scalar multiplication, and vector sum are defined componentwise using the vector-space structure of each Vi.
The direct sum (\oplus_i  V_i  in TeX) is a vector subspace of the product, and consists of all {vi} for which vi=0 for all but finitely many i. This is also the internal sum of the copies of the Vi contained in the product. It coincides with the product if the index set is finite, and if and only if Vi={0} for all but finitely many i.

Q     On the topic of notations not in Axler: what was that alternative notation for the space of polynomials over a field F that Axler denotes by a fancy script P?

A     A[t] is the usual mathematical notation for the set of polynomials in one variable t with coefficients in A. As with vector spaces, one may also say ``over A'' instead of ``with coefficients in A''. The notation does not presume that A is a field -- for instance, Z[t] would be immediately understood as the set of polynomials in t with integer coefficients -- though normally A is at least a commutative ring, in which case so is A[t]. Likewise A[x,y]=A[x][y]=A[y][x] is the ring of polynomials over A in two variables x and y, and similarly A[x,y,z] etc.

Q     I can't find ``quotient spaces'' anywhere in Axler. What were the basic definitions and properties again?

A     Axler does not seem to explicitly use the important notion of a quotient vector space. If U is a subspace of a vector space V, we get an equivalence relation on V by defining two vectors v,v' to be equivalent (``congruent mod U'') if v-v' is in U. The set of equivalence classes then itself becomes a vector space (this must be proved!), called the quotient space V/U. [Note that this is also a notation for V being a vector space over a field U -- we shall strive to make it clear which meaning we intend when both meanings might make sense. At any rate they are pronounced differently: a quotient space is ``V mod U'', whereas a vector space over a field is ``V over F''.]

More generally one can define quotient modules over other rings. For instance the finite cyclic group ``Z/NZ'' really is the quotient of the Z-module Z by its submodule NZ. You might note that Z/NZ itself has the structure of a ring, not just a Z-module. In general, for a commutative ring A, if we consider A as a module over itself, a submodule I is an ``ideal'' (a subgroup closed under multiplication by arbitrary elements of A), and for any ideal I, the quotient A/I inherits the structure of a ring.

Back to quotient vector spaces. If V is finite dimensional, then so is U; by the usual tactic of choosing a basis for U and extending it to a basis for V, we can show that dim(V)=dim(U)+dim(V/U) by proving that the classes of the added basis vectors constitute a basis for V/U. In general dim(V/U) is called the codimension of U in V.

Suppose now that T is a linear transformation from V to some vector space W. We then obtain a well-defined map from the quotient space V/ker(T) to the image of T, and readily check that this map is an isomorphism. Hence the two spaces have the same dimension. Theorem 3.4 now follows without further fiddling with bases.

Q     Suppose V is a finite-dimensional vector space over F. We know now that if F=R then it's easy to tell whether two symmetric bilinear pairings on V are equivalent, and have been told that it's hard if F=Q. What if F is a finite field like Z/pZ?

A     Good question. In this case, it turns out that equivalence is at least as easy to detect as it was over R -- though the proof is still nontrivial.

Let's assume that the characteristic of F, though finite, does not equal 2. (There's a similar answer even in characteristic 2, but it's somewhat harder to state.) Suppose the pairing is nondegenerate -- the general case is easily reduced to this. Choose an orthogonal basis {vi}, and let D be the product of the elements <vi,vi>. Then two nondegenerate symmetric bilinear pairings are equivalent if and only if the ratio of their D's is a square in F. The ``only if'' part is a necessary condition over any field not of characteristic 2; it is easily obtained using determinants (and can be shown even without them when F is finite, for instance by counting solutions of <v,v>=c for each c in F). The ``if'' part requires more thought.

It follows that for each positive integer n there are exactly two equivalence classes of nondegenerate symmetric bilinear pairings on an n-dimensional vector space over F.

Q     What's a ``quadratic form''? I hear this expression in class but I can't find the definition.

A     A ``form of degree d'' is a homogeneous polynomial of degree d in finitely many variables. If these variables are coordinate functions on a vector space V then a form of degree d is an element of the d-th symmetric power of the dual space V*. For d=1,2,3,4,5,6,... one often sees ``of degree d'' written as ``linear'', ``quadratic'', ``cubic'', ``quartic'', ``quintic'', ``sextic'', etc. [You may also encounter the terms ``binary'', ``ternary'', and ``quaternary'', which specify that the number of variables (a.k.a. dim(V)) equals 2, 3, or 4. Nowadays one rarely sees such words for forms in 5 or more variables. I doubt I'll ever have reason even to use the terms binary/ternary/quaternary in Math 55.]

If <·,·> is a symmetric bilinear pairing on V then the map Q from V to F defined by Q(v)=<v,v> is a quadratic form. Conversely, from Q we may recover the pairing using the polarization identity 2<v,w>=Q(v+w)-Q(v)-Q(w), provided F is not of characteristic 2. So, unless 2=0, quadratic forms are equivalent with symmetric bilinear pairings, and the two terms are often used interchangeably. For instance, one may speak of a ``nondegenerate quadratic form'', meaning a quadratic form whose associated pairing is nondegenerate.

Q     Are there any lattices L in Rn that are their own duals and are not isomorphic with Zn?

A     Yes, once n is large enough. The first example is n=8: there's a unique self-dual lattice in R8 not isomorphic with Z8. This is the lattice with the maximal kissing number 240 in that dimension. Can you find it? (The 24-dimensional lattice whose 196560 minimal nonzero vectors attain the 24-dimensional kissing number is also self-dual.)

It is known that for each n there are only finitely many isomorphism classes of self-dual lattices in Rn, but their number grows rapidly as n increases. Here's a table (from SPLAG, page 49):
 n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 ... 32 ... # = 1 1 1 1 1 1 1 2 2 2 2 3 3 4 5 8 9 13 16 28 40 68 117 297 665 ... >108 ...

Q     What's a ``codimension''? I can't find it in the index of Axler.

A     If U is a subspace of a vector space V, the codimension of U in V is the dimension of the quotient space V/U; it is a measure of how far U is from exhausting V. When dim(V) is finite, the codimension also equals dim(V)-dim(U).