How do you prove that $[c,d]$ is not open in $\Bbb R$?

Question

We know that $(a,b)$ are open by definition. How do you prove that some arbitrary union of $(a,b)$ cannot give you $[c,d]$ ?

Answer 1

Because for every union of open intervals, let's call it $U$, we have the property:

$$\forall x\in U \exists\epsilon>0: (x-\epsilon, x+\epsilon) \subseteq U$$

Or, in english,

Each element $x$ of $U$ has some neighborhood $(x-\epsilon, x+\epsilon)$ that is completely included in $U$.

This property does not hold for $c$ and $d$ in the set $[c,d]$.

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The statement above can easily be proven. Let $U$ be some union of intervals,

$$U=\bigcup_{i\in I} (a_i, b_i)$$

(note, $I$ may not be finite or even countable for this proof to work!).

Then, let $x\in U$. By definition of union, there exists some $i\in I$ such that $x\in (a_i, b_i)$. Then, set $\epsilon = \min\{\frac{x-a_i}{2}, \frac{b_i-x}{2}\}$.

We can now prove that $(x-\epsilon, x+\epsilon)\subseteq (a_i, b_i)\subseteq U$:

Let $y\in (x-\epsilon, x+\epsilon)$. Then, we know:

1. $$y>x-\epsilon$$
2. $$x-\epsilon \geq x-\frac{x-a_i}{2} =\frac{x}{2} + \frac{a_i}{2} > \frac{a}{2} + \frac{a}{2} = a $$

So we also know that $y>a_i$.

Similarly, we can show that $y

1. $$y< x+\epsilon$$
2. $$x+\epsilon \leq x+\frac{b_i-x}{2} = \frac{x+b_i}{2}<\frac{b_i+b_i}{2}=b_i$$

so $y

Together, this means that $y\in (a_i, b_i)$ and, consequently, that $(x-\epsilon, x+\epsilon)\subseteq(a_i, b_i)$.

Answer 2

In $\Bbb R$ both open and closed sets are $\Bbb R$ and $\phi$. Now see the complement of $[c,d]$ is $(-\infty ,c) \cup (d, + \infty)$ which is open. Hence $[c,d]$ is closed.