Showing that a half-open set is neither open nor closed

[Mr Davis 97]

Homework Statement

Where $a,b\in \mathbb{R}$, show that $[a,b)$ is not open.

Homework Equations

The Attempt at a Solution

I need to show that there exists an $x\in [a,b)$ such that for all $\epsilon > 0$, $B_\epsilon (x) \not \subseteq [a,b)$. To this end put $x=a$, and let $\epsilon > 0$. Then $B_\epsilon (x)= (a-\epsilon, a+\epsilon)$, and since $a-\epsilon < a$, we have that $B_\epsilon (x) \not \subseteq [a,b)$.

Is noting that $a-\epsilon < a$ enough to prove that one is not a subset of the other?

 

[fresh_42]

Yes.

And why isn't it closed?

 

[WWGD]

Just a comment. I assume you may have meant half-open interval, not half open set since I can't think of any other setting in which a set may be half open.

 

[fresh_42]

Just a comment. I assume you may have meant half-open interval, not half open set since I can't think of any other setting in which a set may be half open.

The set of molecules forming your front door

 

[WWGD]

Actually, I ruined someone's joke that a set is not like a door, open or closed. A door may be open , closed and locked or closed and unlocked. But I can't think of unlocked open sets ;).

 

[fresh_42]

But I can't think of unlocked open sets ;)

"Many of the perpetually open Denny’s restaurants were built without locks, which was problematic when they decided to close down for Christmas for the first time in 1988."