### Problem Set 4

**Closure of the Interior of a Closed Set**
Let A be a closed subset of a topological space. Is it necessarily true
that the closure of the interior of A is equal to A? Either prove that
it is, or produce a counterexample.

**[0,1] in the finite complement topology on R**
Let A=[0,1] in the finite complement topology on R.
Find the closure of A and the
interior of A and informally justify your responses.

**The origin and outside the unit metric ball in R**^{2}.
Let A = {(x,y) in R^{2} | x^{2} +y^{2} > or equal to 1}
U {(0,0)}. Look at the standard topology on A as the subspace topology
in R^{2}.

a) Find the closure of the metric ball or radius 1
about the origin: B( (0,0) , 1)
in Tau_{A} and informally justify your answer.

b) Find the interior of {(0,0)} and informally
justify your answer.

**Which of the following are Hausdorff?**
(Informally justify why or why not.)

a) X={1,2,3} with the topology={Empty set,
{1,2}, {2},{2,3},{1,2,3}}

b) The discrete topology on R

c) The Cantor
Set with the subspace topology induced as a subset of the usual
topology on R

d) Rl, the lower limit topology on R

e) The product topology Rl x R

**Zariski Topology**
Let n be a natural number and let *P* be the collection of
all polynomials in n variables. For p in *P* let

Z(p) = {(x_{1}, x_{2}, ..., x_{n}) in R^{n}
| p((x_{1}, x_{2}, ..., x_{n})) = 0}

ie Z(p) is the set of zeros of the polynomial p.

{Z(p)|p in *P*} is a basis for **the closed sets** of
the Zariski topology on R^{n} (ie the empty set and R^{n}
are closed, the finite union and arbitrary intersection of closed sets
are closed.)

a)
Prove that the the Zariski topology on R^{n} is
T_{1} but not T_{2} (see hints below)

b) Prove that on R, the Zariski topology is the
same as the finite
complement topology by showing that they have the same basis elements.

c) **(Grad)** Prove that on R^{2} the Zariski topology is not
the same as the finite complement topology.

Hints and comments on the Zariski problem:

Do note that for R^{2}, p(x,y) = y is a perfectly fine
polynomial in two variables,
as we send the second variable, x to 0*x, ....
The zeros of this polynomial
would be the set (anything, 0).

You know what a closed set in this topology
look like - the zeros of some polynomial and arbitrary intersections and
finite unions - so that also tells you
what open sets look like (the complement of closed sets).

To show that the Zariski topology
is T_{1}, you can either use the definition,
__or__ prove that a point in R^{n} is
closed (an equivalent formulation
of T_{1} from class). Choose the proof that is easier.

To show that it is not Hausdorff,
what does Hausdorff mean in terms of closed
sets? For all x not equal to y in R^{n}, there exists C, D
closed such that x is not in C, and y is not in D, so that
(C U D) is all of R^{n}. You can negate this and show the negation
holds and so the Zariski topology is not Hausdorff.

The Zariski topology is important in algebraic geometry - there a hyperbolic
paraboloid would be viewed as a closed set in the Zariski topology on
R^{3}. In an abstract algebra II class, one typically examines
the notion of `commutative ring with 1' and of `prime ideal.'
The Zariski topology makes a topological space out of the set of prime
ideals in a commutative ring with 1, by prescribing that the closed sets
should be those subsets of prime ideas which contain a given ideal.
One can make beautiful topological spaces by pasting these together.
One obtains the flexibility of changing the algebraic object
as you move around in your topological space, and the notion of what it
means to be open changes along with it. Some people think that the
universe we live in is a topological space with many dimensions, called
a Calabi-Yau manifold, which can be understood with this type of language.
So even if you are not planning on learning a lot of advanced abstract
mathematics, these abstractions may have everything to do with the
universe we live in.