# Comparing product topologies

1. Oct 27, 2011

### UJS

I'm reading Topology by Munkres, and I'm having some trouble with exercise 5 on page 92 (screenshot attached).

1. The problem statement, all variables and given/known data
Let X and X' denote a single set in the topologies T and T', respectively; let Y and Y' denote a single set in the topologies U and U', respectively. Assume these sets are nonempty.
Show that if T' is finer than T and U' is finer than U, then the product topology on X' x Y' is finer than the product topology on X x Y.

3. The attempt at a solution
I think this is a counterexample...
Let T = T' = U = U' be the standard topology on ℝ. Let X = Y = Y' = (0, 1). Let X' = (2, 3).

X is open in X and Y is open in Y → X x Y is open in X x Y.
X is not a subset of X' →X is not open in X' → X x Y is not open in X' x Y'.
Hence the product topology on X' x Y' is not finer than the product topology on X x Y.

#### Attached Files:

• ###### topology.jpg
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2. Oct 28, 2011

### Kreizhn

So you're dropping to the subset topologies on X,X',Y and Y' right? However, the examples you've chosen are not comparable and so your result is meaningless (check Munkres, he even talks about how some topologies are not comparable). Hence the result only has meaning when you can compare the topologies.

3. Oct 28, 2011

### Deveno

in your example, the topology on X' is not finer than that on X. why? because NONE of the open sets of X (except the empty set) are even in the topology for X'.

remember, a finer topology has to be a superset of the coarser topology, so X' has to have all of the open sets of X, plus more.

the point of the exercise is to consider two differing topology structures on the same underlying space, in other words two different elements of the power set:
P(XxY), given an already-established relationship between two subsets of P(X) and P(Y).