- Mendelson p. 86 #6: Note the typo in the instructions - there Theorem 4.17 should be Theorem 4.11
- Mendelson p. 91 #4
- Prove that X is discrete iff every function f : X-->R is continuous
- Munkres p. 111 # 2
**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**Grad Problem****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) Show that on R, the Zariski topology is the same as the finite complement topology by showing that they have the same basis elements.

b) Give a reason why R^{2}with the Zariski topology is not the same as R^{2}with the finite complement topology.

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

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,prove that a point in the space are closed (an equivalent formulation of T__or___{1}). 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 in the Zarisky topology with x not in C and y 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.