Problem with proof of the upper hemicontinuity of correspondence

Question

I have a problem with a proof I found here of the upper hemicontinuity of the best-reply correspondence in the Nash Theorem.

Below there is the proof, and here my problems:

Problems:

1. Is here implicitly assumed, without loss of generality, that the utility function $u_i$ is increasing in the convergence? Otherwise, I don't see how the red inequality stands.

2. How does the blue inequality stands?
   Is it related to the implicit assumption that I think we have?

Proposition: Let $B_i : \Sigma_{-i} \rightrightarrows \Sigma_i$, defined as $$B_i (\sigma_{-i}) := \arg \max_{x \in \Sigma_{i}} u_i (x , \sigma_{-i}) ,$$ with $B := \prod_{i \in I} B_{i}$. Then, $B$ is upper hemicontinuous.

[Notice that $\Sigma := \Sigma_{i} \times \Sigma_{-i}$ is nonempty compact convex, and the proof relies on the equivalence of upper hemicontinuity to closedness in this context.]

Proof:
By contradiction, assume that $r$ is not closed. Hence, there exists a sequence $(\sigma^n , \hat{\sigma}^n) \to (\sigma , \hat{\sigma})$ with $\hat{\sigma}^n \in B ( \sigma^n)$, but $\hat{\sigma} \notin B ( \sigma)$. If $\hat{\sigma} \notin B ( \sigma)$, then there is some $i \in N$ such that $\hat{\sigma}_i \notin B_i (\sigma)$. Thus, there is a $\sigma^{\prime}_i $ such that $u_i (\sigma^{\prime}_i , \sigma_{-i} ) > u_i (\hat{\sigma}_i , \sigma_{-i} ) + 3\varepsilon$.

From the continuity of $u_i$ and the fact that $\sigma_{-i}^n \to \sigma_{-i}$, for $n$ sufficiently large, we have

$$ \color{red}{u_i (\sigma^{\prime}_i , \sigma^{n}_{-i} ) \geq u_i (\sigma^{\prime}_i , \sigma_{-i} ) - \varepsilon .}$$

Bringing those inequalities together we have

$$\color{blue}{u_i (\sigma^{\prime}_i , \sigma^{n}_{-i} ) > % u_i ( \hat{\sigma}_i , \sigma_{-i} ) + 2 \varepsilon \geq% u_i ( \hat{\sigma}^{n}_i , \sigma^{n}_{-i} ) + \varepsilon ,}$$

where the second inequality comes from the continuity of $u_i$. Hence, this contradicts the assumption that $\hat{\sigma}^{n}_i \in B_{i} ( \sigma^{n}_{-i} )$.

As always, any feedback is greatly appreciated.
Thank you for your time.

Answer 1

This follows from continuity of the (mixed extensions of the) payoff functions alone alone. Since $u_i$ is continuous, $\lim_{n\to\infty}\sigma^{n}_{-i}=\sigma_{-i}$ implies $\lim_{n\to\infty}u_i (\sigma^{\prime}_i , \sigma^{n}_{-i} )=u_i (\sigma^{\prime}_i , \sigma_{-i} )$. Let $\varepsilon>0$. By the definition of convergence, $$u_i (\sigma^{\prime}_i , \sigma^{n}_{-i} )\in \big(u_i (\sigma^{\prime}_i , \sigma_{-i} )-\varepsilon,u_i (\sigma^{\prime}_i , \sigma_{-i} )+\varepsilon\big)$$ for $n$ large enough. For such $n$, we therefore have have $$u_i (\sigma^{\prime}_i , \sigma^{n}_{-i} ) > u_i (\sigma^{\prime}_i , \sigma_{-i} ) - \varepsilon,$$ which proves the red inequality.

A similar argument works for the second blue inequality. Note that there are two blue inequalities.