### 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 R2. Let A = {(x,y) in R2 | x2 +y2 > or equal to 1} U {(0,0)}. Look at the standard topology on A as the subspace topology in R2.
a) Find the closure of the metric ball or radius 1 about the origin: B( (0,0) , 1) in TauA 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) = {(x1, x2, ..., xn) in Rn | p((x1, x2, ..., xn)) = 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 Rn (ie the empty set and Rn are closed, the finite union and arbitrary intersection of closed sets are closed.)
a) Prove that the the Zariski topology on Rn is T1 but not T2 (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 R2 the Zariski topology is not the same as the finite complement topology.

Hints and comments on the Zariski problem:
Do note that for R2, 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 T1, you can either use the definition, or prove that a point in Rn is closed (an equivalent formulation of T1 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 Rn, 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 Rn. 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 R3. 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.