Closure of the intersection of open halfspaces

Question

Suppose $C$ is a nonempty finite intersection of open halfspaces in $\mathbb{R}^n$. Is it true that the closure $C$ is equal to the intersection of the associated closed halfspaces?

I feel this is true, but I don't know how to prove it because the closure of an intersection is not always equal to the intersection of the closure.

Answer 1

As already noted in another answer, the statement does not necessarily hold if the intersection of the open halfspaces is empty (take the halfspaces $x < 0$ and $x > 0$ in $\mathbb{R}$). But it holds if this intersection is non-empty.

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For $\alpha \in \mathbb{R}^n$ with $\alpha \neq 0$ and $c \in \mathbb{R}$ we denote the corresponding open and closed halfspaces by $$ H_{\alpha, c} := \{\lambda \in \mathbb{R}^n \mid (\lambda, \alpha) > c \} \quad\text{and}\quad \overline{H}_{\alpha, c} := \{\lambda \in \mathbb{R}^n \mid (\lambda, \alpha) \geq c \}. $$ Note that $\overline{H}_{\alpha, c}$ is closed. It is in fact the closure of the associated open halfspace $H_{\alpha, c}$: Since $\overline{H}_{\alpha, c}$ is closed and contains $H_{\alpha, c}$, it follows that $\overline{H_{\alpha, c}} \subseteq \overline{H}_{\alpha, c}$. If on the other hand $\lambda \in \overline{H}_{\alpha, c}$, then we can pick some $\mu \in H_{\alpha, c}$ and consider the sequence $$ \lambda_n := \frac{1}{n} \mu + \left( 1 - \frac{1}{n} \right) \lambda. $$ This is a sequence in $H_{\alpha, c}$ because $$ (\lambda_n, \alpha) = \frac{1}{n} \underbrace{(\mu, \alpha)}_{> c} + \left( 1 - \frac{1}{n} \right) \underbrace{(\lambda, \alpha)}_{\geq c} > c. $$ Because $\lambda_n \to \lambda$ we find that $\lambda$ is contained in $\overline{H_{\alpha, c}}$, showing that $\overline{H}_{\alpha, c} \subseteq \overline{H_{\alpha, c}}$.

Now let $\alpha_i \in \mathbb{R}^n$, $\alpha_i \neq 0$ and $c_i \in \mathbb{R}$ for $i \in I$, such that the intersection $\bigcap_{i \in I} H_{\alpha_i, c_i}$ is non-empty. On the one hand we have that $$ \overline{ \bigcap_{i \in I} H_{\alpha_i, c_i} } \subseteq \bigcap_{i \in I} \overline{H_{\alpha_i, c_i}} = \bigcap_{i \in I} \overline{H}_{\alpha_i, c_i}. $$ (The first inclusion is just general point set topology.) If on the other hand $\lambda \in \bigcap_{i \in I} \overline{H}_{\alpha_i, c_i}$, then we can pick some $\mu \in \bigcap_{i \in I} H_{\alpha_i, c_i}$ (where we use the assumption that this set is non-empty) and consider the sequence $$ \lambda_n = \frac{1}{n} \mu + \left( 1 - \frac{1}{n} \right) \lambda. $$ For every $i \in I$ we have that $\mu \in H_{\alpha_i, c_i}$ and $\lambda \in \overline{H}_{\alpha_i, c_i}$, so by our previous argument $(\lambda_n)$ is a sequence in $H_{\alpha_i, c_i}$ for every $i \in I$. Thus we find that $(\lambda_n)$ is a sequence in $\bigcap_{i \in I} H_{\alpha_i, c_i}$. Because $\lambda_n \to \lambda$ we find that $\lambda \in \overline{\bigcap_{i \in I} H_{\alpha_i, c_i}}$, showing that $\bigcap_{i \in I} \overline{H}_{\alpha_i, c_i} \subseteq \overline{\bigcap_{i \in I} H_{\alpha_i, c_i}}$.

Answer 2

In general the statement is not true. E.g. let's consider the two open halfspaces $x < 0$ and $x>0$. Clearly, their intersection is empty and, hence, also its closure $C$. On the other hand, the intersection of the associated closed halfspaces $x \leq 0$ and $x \geq 0$ is $\{0\}\neq C$.

However, I think that the statement is true as long as the intersection of the open halfspaces is not empty, since in this case the intersection of the associated closed halfspaces will be fully dimensional.