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# Cofinite topology; irreducible cover of compact space

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1. Let X be an infinite set, let T be the cofinite topology on X, and let F be the filter generated by the filter base consisting of all the cofinite subsets of X. To which points of X does F converge?

2. Let X be a space. A cover of X is called irreducible if it has no proper subcover.

(a) Prove that X is compact if and only if every open cover of X has an irreducible subcover.

(b) Give an example of a non-compact space X and an open cover of X that has no irreducible subcover.

https://brainmass.com/math/geometry-and-topology/cofinite-topology-irreducible-cover-of-compact-space-126554

#### Solution Preview

1. Let X be an infinite set, let T be the cofinite topology on X, and let F be the filter generated by the filter base consisting of all the cofinite subsets of X. To which points of X does F converge?

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A detailed, step-by-step solution is given in an attached .doc file (Filter&IrreducibleCover-Solutions.doc).

The key points of the proof are (i) that F is the set of all cofinite subsets of X (i.e., that the filter generated by all the cofinite subsets of X is just the set of all cofinite subsets of X) and (ii) that F converges to every ...

#### Solution Summary

The key points in the solutions of the two given problems in topology are provided, and complete, detailed solutions are given in an attached .doc file.

1. Given an infinite set X, the cofinite topology T on X, and the filter F generated by the filter base consisting of all cofinite subsets of X, the set of points of X to which F converges was found.

2. A complete, detailed proof is given of the fact that a space X is compact if and only if every open cover has an irreducible subcover; a counterexample (i.e., an example of a non-compact space X, and an open cover of X that has no irreducible subcover) is also given, together with a justification for the counterexample.

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