Closure of a set is contained in the half space

Question

If $X\subset\mathbb{R}^{d}$ lies in the closed half-space $\mathbb{h}^{+}$, then the closure $\text{cl} X$ of $X$ also lies in $\mathbb{h}^{+}$.

I understand the question but I don't have any idea how to prove this. Any hints/ help would be great. Thanks.

If $A \subseteq B$ and $B$ is closed then $\overline A \subseteq B$. — Kavi Rama Murthy, Mar 13 '21 at 09:50
Oh okay. By $\overline{A}$ i guess you mean $\text{cl} A$? I didn't know this property. — SpeedForce, Mar 13 '21 at 09:55
Not a property, it's the same thing. Just different notation. — Aelx, Mar 13 '21 at 10:23

score 1 · Answer 1 · answered Mar 13 '21 at 10:15

The closure of a set $X$ is (depending on your textbook: by definition) the smallest closed set containing $X$. In fact, such a smallest closed set is guaranteed to exist because the intersection of arbitrarily many closed sets is closed; and that means that the closure is just the intersection of all closed sets containing $X$. In particular, if $h^+$ is one of the closed sets containing $X$, then $h^+$ is one of the sets being intersected to form the closure, i.e., the closure is a subset of $h^+$.

I didn't think it this way. Thanks a lot. – SpeedForce Mar 13 '21 at 10:24 — SpeedForce, Mar 13 '21 at 10:24

Closure of a set is contained in the half space

1 Answers1