Compact domains tame functions

Mathematics · Topology · ages 21-22

Name ______________________ Date ____________

For f(x)=1/x on (0,4], what is the minimum value? Give a decimal.
Answer: ______________
Let K={0} union {1/n:n≥1}. An open-cover member U contains 0. Once U covers all sufficiently late terms, what remains to be covered?
- An uncountable family of isolated points.
- Only finitely many earlier terms 1/n.
- Infinitely many terms bounded away from 0.
- No points, since U must contain all of K.
With the usual distance, which set guarantees that every continuous real-valued function on it is bounded and attains both bounds?
- The open interval (0,1) in R
- The closed interval [0,1] in R
- The rational points of [0,1]
If K is compact and f:K→R is continuous, then f(K) is compact and bounded in R.
Circle one: True False
For f(x)=-x on (0,1], which statement explains why there is no maximum?
- The supremum is 0, but no x in (0,1] gives f(x)=0.
- The maximum is 0, attained at the included endpoint x=1.
- The function is unbounded above because 0 is excluded.
Every closed subset F of a compact metric space X is compact.
Circle one: True False
A metric space X is complete. For every ε>0, finitely many open ε-balls cover X. What follows?
- X is finite, since each ball must contain at most one point.
- X is Euclidean, since every sequence converges in the space.
- X is only bounded; these conditions give no compactness test.
- X is compact, since it is complete and totally bounded.
Which set is closed and bounded in its stated metric space but is not compact?
- The interval [0,1] in R with its usual distance
- A set of five points with any metric on those points
- The closed unit ball in ell^2 with its norm distance
Using distance |x-y|, which statement identifies a set closed and bounded in Q but not compact?
- Q intersect [0,1] is closed in Q but misses irrational limits.
- Q intersect (0,1) is closed in Q and contains its endpoint limits.
- [0,1] in R is closed in Q and contains only rational points.
For an open cover, some δ>0 makes every B(x,δ) lie in a cover member. Nine δ/2-balls cover X. What upper bound does this construction give on the number of original cover members needed?
Answer: ______________

Answer key

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Compact domains tame functions W1-mt_dBLXU0wSN9-s1

0.25 · For 0<x≤4, reciprocals satisfy 1/x≥1/4. Equality holds at the included input x=4, giving minimum 0.25.
Only finitely many earlier terms 1/n. · A neighbourhood of 0 contains the tail of 1/n. Only finitely many earlier terms remain, each covered by one additional chosen member.
The closed interval [0,1] in R · [0,1] is nonempty and compact, so the extreme value theorem applies. On the other sets, continuous real-valued functions can be unbounded.
True · Preimages turn any open cover of f(K) into one of K. A finite selection covers the image. A compact subset of R is bounded.
The supremum is 0, but no x in (0,1] gives f(x)=0. · Every value is negative, while f(1/n)=-1/n approaches 0. Thus 0 is the supremum. Attaining it would require x=0, outside the domain.
True · Extend a relative open cover of F to open sets in X and add X\F. A finite subcover of X leaves finitely many original members covering F.
X is compact, since it is complete and totally bounded. · The small-ball condition is total boundedness. Completeness and total boundedness together characterize metric compactness.
The closed unit ball in ell^2 with its norm distance · The unit vectors in the ell^2 ball remain distance sqrt(2) apart. No subsequence converges. The real interval and finite metric set are compact.
Q intersect [0,1] is closed in Q but misses irrational limits. · Q intersect [0,1] is relatively closed and bounded. Rational approximations to sqrt(2)/2 have no limit in the set, so compactness fails.
9 · Choose a cover member containing the δ-ball about each of the nine centres. These members cover all nine smaller balls and hence X. At most nine are needed.