**A**
Yes: if we know that

**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.

**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. :-)

**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.

**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24) and again on page 170 (in the middle of Exercise 24 of that section), but that definition is either incorrect or outdated, because it does not require that the map be a bijection.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 functionfbetween metric spaces satisfyingd(f(x),f(y)) =d(x,y) for allx,y-- would nowadays be said to be ``an isometry to its image''. Note that such anfis automatically aninjection (why?).

**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.

**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.]

**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} *dx*, *x*=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.

Let {pIs the same thing true in an arbitrary_{n}} be any sequence of elements of E. If there exists p in X such that p_{n}->p then p is in E.

**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 {p_{n}}
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 {p_{n}}, U must contain p_{n}
for all large enough n. But U is disjoint from E,
and therefore can contain no p_{n} 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
{p_{n}} in E such that every open U containing p
also contains all but finitely many p_{n}.
For a metric space we did this using in effect the fact that
the sequence of open sets U_{n}=B_{1/n}(p)
has the the following property:
every open U that contains p must contain some U_{n}.
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 {f_{n}} 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
f_{n}(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 {p_{n}},
replacing the index set {1,2,3,...} by an arbitrary poset
[**p**artially **o**rdered **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 U_{n}=B_{1/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.

**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.

**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 Y^{X}={functions from X to Y} to Y^{E}={functions from E to Y}; then an extension of f is a preimage of f under this map.

**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.

**A**
The **product**
(\prod_i V_i in TeX)
is the usual Cartesian product, and consists of all assignments
{*v _{i}*} of a vector

The

**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.

**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**/*N***Z**''
really is the quotient of
the **Z**-module **Z**
by its submodule *N***Z**.
You might note that **Z**/*N***Z**
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.

**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
{*v _{i}*}, and let

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*.

**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.

**A**
Yes, once *n* is large enough.
The first example is *n*=8:
there's a unique self-dual lattice
in **R**^{8}
not isomorphic with **Z**^{8}.
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 **R**^{n},
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 | ... | >10^{8} |
... |

**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*).