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This is basically because the action spaces P k ( a ) are compact spaces and the utility functions are continuous with respect to the action profile.
(3.20) From [20] and [13], we know that all entries of the matrix operator A are compact, hence, we can easily see that the matrix operator A is compact in the weighted product space Y.
(3.33) From [20] and [13], we know that all entries of the matrix operator A are compact, hence, we can easily see that the matrix operator A is compact in the weighted product space Z.
Note that { x n } ⊂ ⋃ n = 1 + ∞ A n ∪ A, as well as A n and A are compact and A n → A, due to Lemma 2.7, { x n } or its subsequence converges to a point of A. Hence the right-hand side set of (5.1) is well-defined and nonempty.
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(a) is compact ( is relatively compact).
(3) sup ( A ) is compact.
A is compact and continuous; 3.
(2) If A is compact, then A is totally bounded. .
A is compact if and only if A is sequentially compact.
So, ϒ ( A ( B ) ) = 0. Therefore, A is compact.
Now, we prove that A is compact and continuous, respectively.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com