Let (E, r) be a topological space. Let E' = EU {co} with co ¢ E. We define the following topology on E': 9 = (1.1 C E' : (U E r) V (CO E U A CEU is compact)}. • Show that .7 is a indeed a topology. •...

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Answer To: Let (E, r) be a topological space. Let E' = EU {co} with co ¢ E. We define the following topology on...

Robert answered on Dec 22 2021
127 Votes
Exercise-1:
We have
τ = {U ⊂ E′ : (U ∈ τ) ∨ (∞ ∈ U ∧ CEU is compact )}
If U ∈ τ , then we say U i
s of type 1 and if (∞ ∈ U ∧CEU is compact ) then we
say U is of type 2. Second type of set is of the form E′ − C where C is some
compact set in E.
1. (a) The empty set φ ∈ τ and φ ∈ τ and E′ ∈ τ because E′ = E′− φ and
φ is compact.
(b) We will prove that intersection of two open set in open. This will
prove finite intersection of open set is open. We have following three
possibility:
U1 ∩ U2 is of type 1
(E − C1) ∩ (E − C2) = E − (C1 ∩ C2)
which turns out to be of type 2. and third mix of both type of open
set.
U1 ∩ (Y − C1) = U1 ∩ (E − C1)
and this is of type 1.
Hence we see that intersection of finite collection is of either type 1
or type 2.
(c) We have If we have all Uα of type 1, then
∪αUα = U ∈ τ type 1
If Uα are such that
Uα = E
′ − Cα
for Cα compact sets, then we have
∪α(E′ − Cα) = E′ − ∩(Cα) = E′ − C
and E′−∩(Cα) lies...
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