My Question: I want to show that the interval $(0,1)$ is open in $\mathbb{R}$ and that $[0,1]$ is not open in $\mathbb{R}$. I proved the latter, but I felt that it was clumsy and could be refined. If you could also show that $(0,1)$ is open that would be great; these are the simplest problems that I can think of, and I am concerned with the format of the proof (my intuition tells me that they are both true). I am working with the following definitions (I may have made an error while translating the definitions into symbolic logic, if so please correct me). Thanks!
Definitions:
A set $G$
in $\mathbb{R}^{p}$
is said to be open in $\mathbb{R}^{p}$
if for all $x$
in $G$
,there exists an $r>0$
, such that for each $y$
in $\mathbb{R}^{p}$
satisfying the condition $\|x-y\|<r$
also belong to $G$.
Or, even more formally we could write $$G\text{ is an }\textbf{open set} \iff G\subset\mathbb{R}^{p}\wedge\forall x\in G\exists r>0\in\mathbb{R}\left(\forall y\in\mathbb{R}^{p}\left(\|x-y\|<r\iff y\in G\right)\right).$$
Using the definition of an open ball $\mathcal{B}(x,r)=\{y\in\mathbb{R}^{p}\::\:\|x-y\|<r\}$
, where $x$
is the center of the ball and $r>0$
is the radius of the ball, then we can rephrase the definition of an open set by saying that $G$
is open if every point $x$
in $G$
is the center of some open ball contained entirely in $G.$ Again more formally we can write
$$G \text{ is an }\textbf{open set}\iff G\subset\mathbb{R}^{p}\wedge\forall x\in G\exists r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\subset G\right).$$
My work:
Consider the set $F=\{x\in\mathbb{R}\::\:0\leqslant x\leqslant1\}\subset\mathbb{R}$
, is $F$
an open set?
No! proof. Since, we know that $G$
is a subset of $\mathbb{R}^{p}$
, we need only consider the second conjunct in the bi-conditional above. The negation of the statement $\forall x\in G\exists r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\subset G\right)$
is, $\exists x\in G\forall r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\not\subset G\right)$
.
Now let us consider the point $x=1$ in $F$
then we must consider all of the balls centered at $1$ with a radius $r>0$, i.e. ${\cal B}(1,r)=\{y\in\mathbb{R}\::\:|1-y|<r\}$. If we want to show that the negation is false then we need to find a ball such that for each $y\in{\cal B}(1,r)$, $y\in G$. However, by the Archimedes property we have that we know that for all $r>0$
that there will always be a number $y^*=1/n$ such that $0<y^*<r$
, and that $(1+y^*)\in{\cal B}(1,r)$
while $(1+y^*)\notin G$
. Therefore given the element $1$
in $F$
, ${\cal B}(1,r)\not\subset G$
for all $r>0$
; hence, $F$
is not an open set.
Best Answer
Yes, you've got exactly the right idea: Any ball centered at $1$ must contain a number greater than $1$. For a more direct proof that's constructive (as opposed to just invoking the Archimdean property), consider:
As a hint to show that $(0, 1)$ is open, let $x \in (0, 1)$ be given. Consider $$r = \frac{\min(x, 1 - x)}{2}$$ and show that that $\mathcal{B}(x, r) \subseteq (0, 1)$.