PEI Math 521K study materials
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PEI Math 521K: Metric Spaces, Topology of R^n, and Compactness

PEI Math 521K · Work through your exam, one topic at a time.

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Metric Spaces, Topology of R^n, and Compactness

This domain tests open and closed sets, closure and interior, continuity through preimages, and compactness with its characterizations in R^n and in general metric spaces. Distinguishing results that hold only in R^n from those valid in every metric space is a recurring theme.

Metric topology

A metric satisfies positivity, symmetry, and the triangle inequality. Open balls B(x, r) generate the topology. A set U is open if every point of U has a ball contained in U. A set F is closed iff its complement is open iff F contains all its limit points. The closure is the smallest closed set containing S, and the interior is the largest open set inside S. Whether a set is open or closed depends on the ambient space; subspace topology matters.

Continuity and connectedness

f: X → Y is continuous iff the preimage of every open set is open, equivalently iff preimages of closed sets are closed. Continuous images of compact sets are compact, and continuous images of connected sets are connected. In R, the connected sets are exactly the intervals.

Compactness

A set is compact if every open cover has a finite subcover. In metric spaces, compactness is equivalent to sequential compactness, and also to completeness together with total boundedness. Heine–Borel: a subset of R^n is compact iff it is closed and bounded. Continuous functions on compact sets are uniformly continuous and attain extrema. Cantor's intersection theorem: nested nonempty closed subsets of a compact set have nonempty intersection.

Worked example

Show (0, 1] is not compact. Consider U_n = (1/n, 2) for n ≥ 1; each is open in R and their union contains (0, 1]. Any finite subcollection has a largest index N, and its union is U_N = (1/N, 2), which omits every point of (0, 1/N]. So no finite subcover exists, and (0, 1] is not compact, consistent with failing Heine–Borel because it is not closed in R.

Traps

Assuming closed and bounded implies compact in every metric space; it fails for the closed unit ball of C[0, 1] with the sup norm. Confusing not open with closed. Forgetting to specify the ambient space. Verifying sequential compactness for only one sequence instead of all sequences. Applying Heine–Borel to a set that is not a subset of some R^n.