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Consider the set of points $$O = \{ x \in P \mid \alpha^* = C^T x \}$$ where $P \subseteq \mathbb R^n$ is a closed convex set, $C \in \mathbb R^n$ and $\alpha^* = \min \{ C^Tx \}$. Then, $O$ is closed convex set.
This seems a pretty simple statement in my linear programming class but I am unsure how to show it formally. I can easily show it is a convex set but I am not sure how to show it is a closed set.
If $x_k \in O$ and $x_k \to x$ then $C^{T}x_k \to C^{T}x$ and $\alpha^{*}=C^{T}x_k$ for each $k$. Hence $\alpha^{*}=C^{T}x$ and $x \in O$.
You have $$ O = P \cap \{x \in \mathbb R^n | \alpha^* = C^\top x\},$$ i.e., it is an intersection of two closed sets. Hence it is closed.
First, the function $f(x)=C^T x$ is a finite-dimensional linear function, and therefore continuous.
Also, in $\mathbb R^n$ single-element subsets are always closed; $\{\alpha^*\}$ is such a set.
Now the preimage of closed sets under continuous functions is closed. The preimage of $\{\alpha^*\}$ is $f^{-1}[\{\alpha^*\}]=\{x\in\mathbb R^n|C^Tx=\alpha^*\}$. Therefore this set also is closed.
Finally, $P$ is closed by assumption, and the intersection of two closed sets is closed. But $P\cap f^{-1}[\{\alpha^*\}] = \{x\in P|C^Tx=\alpha^*\}$, which is exactly the set you asked about.
