How to show that this set is closed?

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.

• Why are some vectors uppercase and others lowercase? Oct 29 '19 at 6:08
• It was given like this in the lecture slides. I guess C is uppercase as it is constant and x is not any particular vector. Oct 29 '19 at 6:24
• Whoever wrote the lecture slides has no taste. Good notation provides a "type system". Oct 29 '19 at 6:25

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.