Download Topology - Homework Sets 8 and 9

Topology - Homework Sets 8 and 9
Due Tuesday, April 5
1. Determine which of the following subsets of R2 are compact.
(i) ( x, y) ∈ R2 | ( x/3)2 + (y/5)2 = 1
This is compact. To see this, recall that for Rn , a subset is compact if
and only if it is closed and bounded. This is the zero set of a continuous
function, and so it is closed (see problem 2 below for more details of this
type of argument). It is also bounded, since it is contained in the ball of
radius 5 centered at the origin.
(ii)
( x, y) ∈ R2 | ( x/3)2 − (y/5)2 = 1
This is not compact. To see this, recall that for Rn , a subset is compact
if and
ponly if it is sequentially compact. Consider the sequence of points
(n, 5 (n/3)2 − 1) for n ∈ Z+ . This lies in the above set and has not
convergent subsequence (the x-coordinates diverge to +∞).
(iii)
( x, y) ∈ R2 | x, y ∈ Q, and | x | + |y| ≤ 1
This is not compact since it is not closed: The point (1/π, 1/π ) is a limit
point of this set that is not contained in this set.
2. Define the 2-sphere to be the subspace
S2 := ( x, y, z) ∈ R3 x2 + y2 + z2 = 1 .
Show that S2 is compact.
Since S2 is a subset of R3 , it suffices to show that S2 is closed and bounded.
It is obviously bounded since it is contained in the ball of radius 2 centered at
the origin. To see that it is closed, consider the function f : R3 → R defined by
f ( x, y, z) = x2 + y2 + z2 .
This is a continuous function, so the inverse image of the closed set {1} is
closed. This shows S2 is closed since
f −1 (1 ) = S 2 .
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3. Suppose X is a topological space, and
C1 , . . . , Cn ⊂ X
are compact subspaces. Show that the union ∪nj=1 Cj is compact.
4. Suppose X is a Hausdorff space, and let A, B ⊂ X be disjoint compact
subspaces. Show that there exist disjoint open sets U and V such that
A ⊂ U,
and
B ⊂ V.
5. Let X be a topological space, E ⊂ X a compact subset, and F ⊂ X a closed
subset. Prove that E ∩ F is compact. (Note: In this question it is not assumed
that X is Hausdorff.)
6. Show that sequential compactness (see Munkres p.179) is a topological
property. That is, show that if X is sequentially compact and h : X → Y is
a homeomorphism, then Y is sequentially compact.
7. Consider R2 with the Zariski topology. For each of the following subsets
of R2 , determine whether it is open and whether it is closed.
(i) ( x, y) ∈ R2 | ( x/3)2 − (y/5)2 = 1
This is closed since it is the zero set of the polynomial ( x/3)2 − (y/5)2 −
1. It is not open, since otherwise it and its complement would be a separation of R2 , but no separation exists since R2 is connected in the Zariski
topology.
(ii) x-axis ∪ y-axis
This is closed since it is the zero set of the polynomial xy. It is not open
by the same reasoning as (i).
(iii) R2 − {(0, 0)}
This is open since its complement is the zero set of the polynomial x2 + y2 .
It is not closed by the same reasoning as (i).
(iv) Q × {0}
This is neither open nor closed.
Proof 1: One way to see this is to use the fact that the Zariski topology
is coarser than the standard topology. The set Q × {0} is neither open
nor closed in the standard topology, so the same is true of the Zariski
topology.
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Proof 2: Suppose Q × {0} is closed. Then it is the zero set of a finite
number of polynomials p1 , . . . , pK . Note that (π, 0) is not in Q × {0}.
However, there is a sequence ( xn )n of rational numbers with the property
that xn converges to π in the standard topology (e.g., take xn to be the n
place decimal expansion of π). Then ( xn , 0) ∈ Q × {0}, so
f k ( xn , 0) = 0, ∀1 ≤ k ≤ K, ∀n ∈ Z.
Since polynomials are continuous (in the standard topology), for each
1 ≤ k ≤ K we can take the limit in n to get
f k (π, 0) = 0.
This implies (π, 0) ∈ Q × {0}, which is a contradiction. Hence Q × {0}
is not closed.
A similar argument shows it is not open.
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