Resolvability is open subspace-closed

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This article gives the statement, and possibly proof, of a topological space property (i.e., resolvable space) satisfying a topological space metaproperty (i.e., open subspace-closed property of topological spaces)
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Statement

Suppose X is a resolvable space and U is an open subset of X. Then, U is a resolvable space with the subspace topology.

Facts used

  1. Intersection of dense subset with open subset is dense in the open subset

Proof

Given: A topological space X with disjoint dense subsets C and D. An open subset U of X.

To prove: U has two disjoint dense subsets.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 C \cap U and D \cap U are both dense in U. Fact (1) C,D dense in X
U open in X
given-fact direct
2 C \cap U and D \cap U are disjoint. C,D are disjoint basic set theory!
3 C \cap U and D \cap U are the desired disjoint dense subsets in U Steps (1), (2) Step-combination direct