Download Chapter 4 Compact Topological Spaces

Proof. ⇒ Assume that x is an accumulation point of (xα )α∈J . By the definition of
accumulation point of a net we have for each open set U containing x
KU = {α ∈ J : xα ∈ U } is cofinal in J.
(4.9)
Let K = {(α, U ) ∈ J × Nx : xα ∈ U }, where Nx is the collection of all open sets
containing x. From Eq. (4.9), KU 6= φ. (Fix α ∈ J. Now KU is cofinal in J implies
there exists β ∈ KU such that β ≥ α.) For (α, U ), (β, V ) ∈ K define (α, U ) ≤ (β, V )
if and only if α ≤ β and V ⊆ U (reverse set inclusion). It is easy to see that (K, ≤)
is a directed set. It is given that (xα )α∈J is a net in X. Hence (J, ≤) is a directed set
and f : J → X is such that f (α) = xα . Now define g : K → J as g(α, U ) = α (refer
Eq. (4.9)).
Claim: g(K) is cofinal in J.
So take α ∈ J. Now KU is cofinal in J (refer Eq. (4.9)) there exists β ∈ KU
such that β ≥ α. Now β ∈ KU implies xβ ∈ U that is (β, U ) ∈ K is such that
g(β, U ) = β ≥ α implies g(K) is cofinal in J. Also (α, U ), (β, V ) ∈ K, (α, U ) ≤ (β, V )
implies g(α, U ) = α ≤ β = g(β, V ). Hence f ◦ g : K → X is a subnet of f (or say
f (α) = (xα )). Now let us prove that this subnet converges to x. So take an open set
U containing x. This implies KU is cofinal in J. Fix (α0 , U ) ∈ K. Now α0 ∈ J, KU
is cofinal in J implies β0 ∈ KU such that β0 ≥ α0 . Hence (α, V ) ∈ K, (α, V ) ≥
(α0 , U ) implies (f ◦ g)(α, V ) = f (α) = xα ∈ V ⊆ U . That is for each open set U
containing x there exists (α0 , U ) ∈ K such that (α, V ) ∈ K (α, V ) ≥ (α0 , U ) implies
(f ◦ g)(α, V ) ∈ U . This proves that f ◦ g → x.
Conversely, suppose there is a subnet of (f (α))α∈J = (xα )α∈J which converge
to an element x ∈ X. A subnet of f converges to x means there exists a directed set
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