Another topology on the naturals

Thread starter radou
Start date Aug 27, 2010
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Topology

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SUMMARY

The discussion focuses on the topology defined on the set X = {1, 2, 3, 4, 5} with a subbasis V = {{2, 3, 4}, {1, 4, 5}, {2, 4, 5}, {1, 3}}. It concludes that the dense subset D = {1, 3, 4} satisfies the condition that its closure Cl(D) equals X. Additionally, the mapping f : (X, U) --> (X, P(X)), defined by f(x) = x, is confirmed to be continuous as the preimage of every open set in the codomain is open in the domain.

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Understanding of basic topology concepts, including dense subsets and closures.
Familiarity with subbases and bases in topological spaces.
Knowledge of continuous mappings between topological spaces.
Comprehension of power sets, specifically P(X).

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Study the properties of dense subsets in topological spaces.
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Aug 27, 2010

radou

Homework Helper

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.