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Transcript
closed set in a subspace∗
yark†
2013-03-21 19:47:48
In the following, let X be a topological space.
Theorem 1. Suppose Y ⊆ X is equipped with the subspace topology, and A ⊆
Y . Then A is closed in Y if and only if A = Y ∩ J for some closed set J ⊆ X.
Proof. If A is closed in Y , then Y \ A is open in Y , and by the definition of the
subspace topology, Y \ A = Y ∩ U for some open U ⊆ X. Using properties of
the set difference, we obtain
A
= Y \ (Y \ A)
= Y \ (Y ∩ U )
= Y \U
= Y ∩ U {.
On the other hand, if A = Y ∩ J for some closed J ⊆ X, then Y \ A =
Y \ (Y ∩ J) = Y ∩ J { , and so Y \ A is open in Y , and therefore A is closed in
Y.
Theorem 2. Suppose X is a topological space, C ⊆ X is a closed set equipped
with the subspace topology, and A ⊆ C is closed in C. Then A is closed in X.
Proof. This follows from the previous theorem: since A is closed in C, we have
A = C ∩ J for some closed set J ⊆ X, and A is closed in X.
∗ hClosedSetInASubspacei
created: h2013-03-21i by: hyarki version: h37460i Privacy
setting: h1i hTheoremi h54B05i
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You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
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