half-space: either an open half-space, a closed half-space, or a set $H$ s.t. $$ H^{o} \subsetneq H \subsetneq \bar{H} \,,$$ where the relative interior $H^o$ is an open half-space and the relative closure $\bar{H}$ is a closed half-space. In other words it is an either an open half-space or a closed half-space "modulo the relative boundary".
As defined above half-spaces don't have to be convex (see the community wiki below), so the claim for which I am seeking a counterexample is:
Claim: Every convex set is the intersection of half-spaces (as defined above).
Most of the related questions on this website (e.g. 1 2 3), as far as I can tell, only answer this question for the case of closed convex sets. However, for my purposes it does not matter at all whether or not the convex set is closed. Thus, requiring the intersecting half-spaces to be closed is too restrictive.
The accepted answer to another related question comes closest to addressing this question. However, the counterexample is not valid using the definition of half-space given above, since one can use as one of the intersecting half-spaces $$\{ (x,y): y > 1 \text{ or } (y=1 \text{ and }x < 1) \} \,.$$
The reason why I suspect the claim is false is because it seems to me like such a simple/natural characterization of arbitrary convex sets that, if the characterization were true, one would see it in a textbook somewhere.
The answer to this other related question may implicitly give the answer to the question, since it seems to address why the convex set $C$ would still be the intersection of half spaces even when $\operatorname{relbd}(C) \setminus C \not= \emptyset$.
Best Answer
Let $C \subset \mathbb R^n$ be convex. It is clear that you can write $$ \bar C = \bigcap_{i \in I} H_i $$ for some index set $I$ and some closed half-spaces $H_i$. Now, for every $x \in \mathrm{bd}(C) \setminus C$, there is a closed half-space $\tilde H_x$ with $C \subset \tilde H_x$ and $x \in \mathrm{bd}(\tilde H_x)$. Then, $H_x := \tilde H_x \setminus \{x\}$ is a half-space and we have $$ C = \bigcap_{i \in I} H_i \cap \bigcap_{x \in \mathrm{bd}(C)\setminus C} H_x.$$