But in that moment, I mostly wondered where I was and how long had passed since I last closed my eyes.
For each i ∈ I, since K i is closed, it follows that K is closed and x ∗ ∈ K. Now, { x n } ⊂ S 1 implies that, for each i ∈ I, there exists y i n ∈ T i 1 ( x n ) such that x i n ∈ A i 1 ( x n ), 0 ∈ G i 1 ( x n, y n, z i ). for all z i ∈ A i 1 ( x n ).
For any i ∈ {1,..., N}, there exists a subsequence of (x n ) which belongs to C i. Since C i is closed, we conclude that c ∈ C i, for any i ∈ {1,..., N}.
Since A i is closed for all i = 1, 2,..., k, we conclude ν ∈ ∪ i = 1 k A i, and also we conclude that ∩ i = 1 k A i ≠ ϕ.
Since G i is closed and e i is continuous, we obtain 0 ∈ G i ( p ¯, x ¯, y ¯, z ¯ i ) + B + ( 0, ϵ ) e i ( x ¯ ). for all z ¯ i ∈ A i ( x ¯ ).
Since Graph ( F i ) is closed in X i × X, ( y i V, x V ) ∉ Graph ( F i ), which implies that y i V ∈ G V i ( x V ) and x V ∉ F i ( y i V ).
Since each A i is closed for i ∈ { 1, …, m }, we get that y ∈ ⋂ i = 1 m A i. Then ⋂ i = 1 m A i ≠ ∅ and we can consider the restriction.
Since each A i is closed, we deduce that z ∈ ⋂ i = 1 m A i.
(The inquiry has since been closed).
The school has since been closed.
