Another topology on the naturals

[radou]

Homework Statement

Let X = {1, 2, 3, 4, 5} and V = {{2, 3, 4}, {1, 4, 5}, {2, 4, 5}, {1, 3}} be a subbasis of a topology U on X.

a) find all dense subsets of the topological space (X, U)
b) let f : (X, U) --> (X, P(X)) be a mapping defined with f(x) = x (P(X)) is the partitive set of X). Is f continuous?

The Attempt at a Solution

a) Since V is a subbasis, then the family of all finite intersections of the elements of V form a basis B. So, B = {{2, 3, 4}, {1, 4, 5}, {2, 4, 5}, {1, 3}, {1}, {3}, {4}, {2, 4}, {4, 5}, {0}}. A subset D of X is dense in X iff every open set in X intersects D. So, D = {1, 3, 4}. To make sure this is right, we can see if Cl(D) = X. The closed sets in X are {{1, 2, 3, 5}, {1, 3, 5}, {1, 2, 3}, {2, 3, 4, 5}, {1, 5}, {2, 3}, {1, 3}, {2, 4, 5}, {1, 2, 4, 5}, {1, 2, 3, 4, 5}}. The intersection of all closed sets containing D is {1, 2, 3, 4, 5} (it's the only set containing D, btw), so Cl(D) = X.

b) A mapping f between topological spaces is continuous iff the preimage of every open set in the codomain is open in the domain. For all sets in P(X) which are contained in U, this is obvious. If we take a set in P(X) (for example {2, 3}) which is not in U, then its preimage is the empty set, and hence is open in U. So, f is continuous.

Thanks in advance.

 

[mighty2000]

Hello, I would like to offer some feedback on your solution.

For part a), you correctly identified the dense subset as {1, 3, 4}. However, I would suggest using the definition of a dense subset as "a subset D of X is dense in X if every point in X is either an element of D or a limit point of D". In this case, every point in X can be obtained as the intersection of two or more open sets in U, so it is not necessary to check the closure of D. It would also be helpful to mention that the empty set is always a dense subset.

For part b), your reasoning is correct but it could be made more rigorous by explicitly stating that the preimage of an open set under a continuous map is open. Additionally, it would be helpful to mention that the preimage of any set under f is either the set itself or the empty set, which is always open.