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TOPOLOGY HW 5
CLAY SHONKWILER
28.2
Show that [0,1] is not limit point compact as a subspace of Rl.
Proof. Define the infinite subset A = {1 − 1/k|k ∈ N} ⊂ [0,1]. We want
to demonstrate that A has no limit points in Rl. First, we show that no
element of A is a limit point of A. To see this, let n ∈ N. Then 1−1/n ∈ A.
1 − 1/n ∈[1 −1n,1 −1n + 1),which is an open set in [0,1] as a subspace ofRl. Since
[1 − 1n,1 − 1n+1)intersects A only at 1 − 1n, we see that 1 − 1/n is not a limit point of A.
Now, let x ∈ (0,1),x /∈ A. Then x ∈[1 − 1j,1 − 1j+1)= Ux for some j ∈ N. Since Ux intersects A only at 1 − 1/j, we see that the open set[x,1 − 1j+1)x does not intersect A. Hence, x is not a limit point of A.
We only need to show, then, that 1 is not a limit point of A. But thisis clear, as[1,2) is open in Rl and [1,2) ∩ [0,1] = {1}. Therefore, {1} is aneighborhood of 1 which certainly does not intersect A, so 1 is not a limitpoint of A.Since A is an infinite subset of [0,1] with no limit points, we see that [0,1]
is not limit point compact as a subspace of Rl.
D
28.3
Let X be limit point compact.
(a) If f : X → Y is continuous, does it follow that f(X) is limit point
compact?Answer: Yes. Let A be an infinite subset of f(X). Then A = {f(x)|x ∈
B} where B ⊆ X is infinite. Since X is limit point compact, B has a limit
point b. Let Vb be a neighborhood of f(b). Then, since f is continuous,
there exists some neighborhood Ub of b such that f(Ub) ⊆ Vb. Since b is a
limit point of B, there exists some y ∈ B such that y = b, y ∈ Ub. Thus,
f(y) ∈ f(Ub) ⊆ Vb,
so, since f(y)∈ A, Vb intersects A at some point other than f(b). Since
our choice of neighborhood for f(b) was arbitrary, we conclude that every
neighborhood of f(b) intersects A somewhere other than f(b), meaning f(b)
1
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CLAY SHONKWILER
is a limit point of A. Since our choice of infinite subset A was arbitrary,
we conclude that every infinite subsetof f(X) has a limit point, so f(X) is
limit point compact.

(b) If A is a closed subset of X, does it follow that A is limit point
compact?
Answer: Yes. Let Y ⊆ A ⊆ X be infinite. Then, since X is limit
point compact, Y has a limit point x ∈ X. That is to say that any open
neighborhood of x intersects Y at some point y = x. Since y ∈ Y ⊆ A, x
is a limit point of A. However, since A isclosed in X, A contains all of its
limit points, so x ∈ A. Hence, Y has a limit point in A. Since our choice of
infinite set Y was arbitrary, we conclude that every infinite subset of A has
a limit point in A, so A is limit point compact.

(c) If X is a subspace of the Hausdorff space Z, does it follow that X is
closed in Z?
Answer: No. By Theorem 17.11, SΩ is Hausdorff in the order topologyand, as we saw in Example 2, SΩ is limit point point compact in SΩ. How-
ever, SΩ is not closed in SΩ since it does not contain Ω, which is a limit
point of SΩ.

29.1
Show that the rationals Q are not locally compact.
Proof. Suppose Q is locally compact. Then there exists a compact subspace
of Q containing a basic neighborhood (a, b) ∩ Q of 0. Let x be an irrational
element of (a, b).Then there exists a cauchy sequence {xj} converging to x
such that each xj ∈ Q ∩ (a, b). Then the only limit point of the sequence
of xj’s is x. However, x /∈ Q, so {xj}, when viewed as a set, has no limit
points in Q and, therefore, none in C. Hence, C is not limit point compact.
Since Q is hausdorff, this means C is not compact. From this contradiction,
we conclude that Q is not, in fact,locally compact.
D
29.3
Let X be a locally compact space. If f : X → Y is continuous, does
it follow that f(X) is locally compact? What if f is both continuous and
open?
If f is both continuous and open, then f(X) is locally compact. To see
this, let y ∈ f(X). Then f−1(y) = x ∈ X. So, since X is locally compact,
there exists a compact subspace C of X that contains a neighborhood U of
x....
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