Game theory: proof V is convex and compact Consider a non-cooperative game $(N,S_i,u_i)$.
$\bullet \ N = \{1,...,n\}$ is the set of players.
$\bullet$ For every player $i$, the set $S_i$ is the finite set of pure strategies.
$\bullet$ For every player $i$, the function $u_i: (S_1 \times ... \times S_n) \to \mathbb{R}$ is the utility function of $i$.
A belief of player $i$ about the opponents' choices is a probability distribution $\beta_i \in \Delta S_{(-i)}$, with: $$S_{-i} := S_1 \times ... \times S_{i-1} \times S_{i+1} \times ... \times S_n$$
Suppose that player $i$ has belief $\beta_i$ about the opponents' choices. Then, for every pure strategy $s_i$ he can choose, the corresponding expected utility would be given by: $$U_i(s_i,\beta_i) := \sum_{s_{-i} \in S_{-i}} \beta_i (s_{-i}) u_i(s_i,s_{-i})$$
The strategy $\hat{s_i}$ is called optimal under the belief $\beta_i$ if: $$U_i(\hat{s_i},\beta_i) \geq U_i(s_i,\beta_i) \ \forall s_i \in S_i$$
The strategy $s_i$ is called rational if there is some belief $\beta_i$ about the opponents' choices under which $s_i$ is optimal. For a given belief $\beta_i$ this boils down to whether the believed utility difference vector $v(\hat{s_i},\beta_i)$ is non-negative where: $$v(\hat{s_i},\beta_i)=[U_i(\hat{s_i},\beta_i) - U_i(s_i,\beta_i)]_{s_i \in S_i} \in \mathbb{R}^{|S_i|}$$
Let this be the set of all believed utility difference vectors: $$V_i(\hat{s_i})=\{v(\hat{s_i},\beta_i)\ |\ \beta_i \in \Delta (S_i)\}$$

Is there anyone here who can help me with proving that $V_i(\hat{s_i}$ is convex and compact? So far we've only had a introductory course in game theory, mostly about nash equilibria and determining strategies from matrices... Im quite overwhelmed by the first question of our follow-up course...
This is all I know:
A set is compact $\leftrightarrow$ every sequence in the set has a convergent subsequence which converges to a element in that set.
We can assume $\Delta (S_{-i})$ is compact.
So, actually we only have to prove that a continuous function transforms a compact domain into a compact image, but how?
Anything would be appreciated!!
----EDIT----
Is this enough for convexity:
$V_i(\hat{s_i})$ is convex if for any $v(\hat{s_i},\beta_i),w(\hat{s_i},\beta_i) \in V_i(\hat{s_i})$ the point $t \cdot v + (1-t) \cdot w \in V_i \ \forall \ t \in [0,1]$.
$t \cdot v(\hat{s_i},\beta_i) + (1-t) \cdot w(\hat{s_i},\beta_i)$
=>
$t \cdot [U_i(\hat{s_i},\beta_i) - U_i(s_i,\beta_i)] + (1-t) \cdot [U_i(\hat{s_i},\beta_i) - U_i(s_i,\beta_i)]$
=> $[U_i(\hat{s_i},\beta_i) - U_i(s_i,\beta_i)]= a$
$t \cdot a + (1-t) \cdot a = a \in V_i$
 A: The proof of convexity in the Edit does not make sense because of the following:
When we pick $v$ and $w$ in $V_i$ and consider their convex combination, the crucial detail to pay attention is that if $v\neq w$ then their respective beliefs are different, say $v=v_i(\hat{s}_i,\hat \beta_i)$ and $w=v_i(\hat{s}_i,\tilde \beta_i)$.
Think about beliefs as the indexes of the elements of $V_i$. We know that payoffs are linear in probabilities (beliefs) so $$\alpha v + (1-\alpha) w = v_i(\hat{s}_i,\alpha\hat\beta_i+(1-\alpha)\tilde \beta_i)$$
The proof of convexity is almost done.
As for compactness: you are right, a continuous function maps compact sets into compact sets. Assume the image is not compact. Then you have some sequence in the image that is "escaping": $y_n \in f(X)$ but $\lim y_n \not\in f(X)$. Let $x_n\in f^{-1}(y_n)$, the $x_n$ has a subsequent that converges $\lim x_{n_k} \in X$ but $f(x_{n_k})$ is also converging to $\lim y_n$. Now use continuity to prove $f(\lim x_{n_k})=\lim y_n$. Do you see the contradiction?
