Prove that closed interval is not open

Question

"Prove that the set $A=[a,b)$ is not open." I am looking for a rigorous proof.

My approach:

Suppose that $A$ is actually open, which, by definition, would mean that for any $x\in A$ there would exist some $\epsilon>0$ such that the neighbourhood of $x$: $N_\epsilon(x)=(x-\epsilon,x+\epsilon)\subseteq A$.

Choose $x=a$, then we have $N_\epsilon(a)=(a-\epsilon,a+\epsilon)$. Now, we find a point $\xi \in N_\epsilon(a)$ such that $\xi\notin A$. For all $\xi\in N_\epsilon(a)$ we have $a-\epsilon<\xi<a+\epsilon$. We choose $\xi=a-\epsilon/2$ therefore we have $a-\epsilon<\xi<a<a+\epsilon$. By definition of interval, $A=\{x|x\geq a \wedge x<b\}$. That shows that $\xi\notin A$ therefore $N_\epsilon(a)\not\subseteq A$ for any $\epsilon>0$.

My question is: Is this proof valid and rigorous enough? Can it be made simpler?

Answer 1

It looks very good. I would recommend a slight adjustment, though. You say the following:

For all $\xi\in N_\epsilon(a)$ we have $a-\epsilon<\xi<a+\epsilon$.

This is true, but not used in your proof. Rather, you use the converse of this statement, so I'd say the following, instead:

For all $\xi$ such that $a-\epsilon<\xi<a+\epsilon,$ we have $\xi\in N_\epsilon(a).$

Then, you have by $a-\epsilon<\xi<a<a+\epsilon$ that $\xi\in N_\epsilon(a)$ and $\xi\notin A,$ by reasoning as you did from there.

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Aside from the observation above, there is one more issue that should be clarified for rigor: it isn't clear where the $\epsilon$ comes from in "...then we have $N_\epsilon(a)=(a-\epsilon,a+\epsilon).$" It seems that you were going for a proof by contradiction, so you probably meant something like this:

...then since $a\in A,$ we have $N_\epsilon(a)\subseteq A$ for some $\epsilon>0$ by our assumption that $A$ is open.

From there with those few tweaks, since you later prove that $N_\epsilon(a)\nsubseteq A,$ you'll have completed a proof by contradiction. However, you didn't actually need to proceed that way, as you're very close to a direct proof that $A$ is not open, by demonstrating that there is an $x\in A$ such that for all $\epsilon>0,$ we have $N_\epsilon(x):=(x-\epsilon,x+\epsilon)\nsubseteq A.$ This is because "$A$ is not open" is the formal negation of "$A$ is open," which looks like any of the following equivalent formal statements: $$\neg\bigl(\forall x\in A, \exists\epsilon>0:N_\epsilon(x)\subseteq A\bigr)\\\exists x\in A:\forall \epsilon>0, \neg\bigl(N_\epsilon(x)\subseteq A\bigr)\\\exists x\in A:\forall \epsilon>0, N_\epsilon(x)\nsubseteq A.$$ If you go the direct route, then instead, you'd want to start something like this:

Take any $\epsilon>0$ and consider $N_\epsilon(a):=(a-\epsilon,a+\epsilon).$ We show that $N_\epsilon(a)\nsubseteq A,$ which, since $a\in A,$ will show that $A$ is not open. To do so, we will find a point $\xi\in N_\epsilon(a)$ such that $\xi\notin A.$

From there, you can proceed in much the same way, with the minor tweak I suggested above.