In an attempt to self-study, I am trying to answer the following question:
Let X be an Euclidean space. Suppose that $a \in X\setminus \{ 0 \}$ and let $\beta \in \mathbb{R}$. When is $A = \{x \in X \mid \langle x, a \rangle \neq \beta\}$ convex?
I noticed that if I choose $x, y$ such that $\langle x, a \rangle = \beta + \epsilon$, and $\langle y, a \rangle = \beta - \epsilon$, then if I choose $\lambda = \frac{1}{2}$, I would get: $$\langle \lambda x + (1 - \lambda) y, a \rangle = \langle \lambda x, a\rangle + \langle (1-\lambda) y , a\rangle= \lambda \langle x, a\rangle + (1-\lambda)\langle y, a\rangle = \lambda (\beta + \epsilon) + (1-\lambda) (\beta - \epsilon) = \beta$$
So $x \in A$ and $y \in A$, but $(\frac{1}{2} x + \frac{1}{2} y) \not\in A$.
I then concluded that if there is no $x\in A, y\in A$ such that $\langle x+y, a\rangle = 2 \beta$, the set should be convex, however, I have no idea if my answer is correct, so I asked it here.
Disclaimer: I searched the website and did not find any similar questions but if there are any, please mention them in the comments and I will delete my question.