Proving whether the sets are open or closed in $\mathbb{R}$

Question

I've recently started studying topology and struggling with the concepts. I'd greatly appreciate if anyone can verify, as well as correct my proof, or concept should my proof be wrong.

Question: Prove whether the following sets are open or closed in $\mathbb{R}$

1){${G:={x \in \mathbb{R} | 0<x<1}}$}

Claim: the set is open

Proof: $x \in G$ iff $ x \in (0,1)$. Then, take $\epsilon = \frac{x-0}{2}$ for $x \leq 0.5$ and take $\epsilon = \frac{|x-1|}{2}$ for $x > 0.5$. Then, for any $x \in G$, $(x-\epsilon, x+\epsilon) \subset G$. Then, $ G $ is open in $\mathbb{R}$.

2) Prove that any open interval $I:= (a,b)$ is open. Again, using the aforementioned proof, I can choose my epsilon to be the smaller of $\frac{x-a}{2}$ or $\frac{|x-b|}{2}$ and the claim follows.

3) $I=[0,1]$ is not open. There does not exist any $\epsilon>0$ such that for $0,1 \in I$, $V_{(\epsilon)} \subset I$. I don't know if this is a valid proof. It sounds as if I'm stating something instead of actually proving it. Any leads?

4) $I=[0,1]$ is closed. We have to prove that $\mathbb{R}-I = (-\infty, 0) \cup (1, \infty)$ is open. Can anyone please help in proving this?

Thank you.

Answer 1

1) It's OK if you prove the inclusion $(x-\varepsilon , x+\varepsilon)\subset G$.

3) Simply note that $\forall \varepsilon>0$, $-\dfrac{\varepsilon}{2}\notin [0;1]$ but $-\dfrac{\varepsilon}{2}\in (-\varepsilon, \varepsilon)$ thus $(-\varepsilon, \varepsilon)\not\subset [0;1]$. Similarly, $1+\dfrac{\varepsilon}{2}\in(1-\varepsilon , 1+\varepsilon)$ but not in $[0;1]$.

4) Prove that $(-\infty,0)$ and $(1;+\infty)$ are open as before and use the fact that a union of open sets is open.

Answer 2

hint

For the fourth.

Let $(x_n) $ be a convergent sequence of elements in $[0,1] $.

$$\forall n \;\; 0\le x_n \le 1$$

$$\implies 0\le \lim_{+\infty}x_n\le 1$$

$$\implies \lim_{+\infty}x_n\in [0,1] $$

$$\implies [0,1] \text { is closed} .$$

This is not true for an open set. Take $x_n=\frac {1}{n}\in (0,1) $

but $\lim_{+\infty}x_n=0\notin (0,1) $.

Answer 3

3) Suppose its open and get to a contradiction by using the middle point.

4) Prove that $(-\infty, 0)$ is open and that $(0, \infty)$ is open so $R\setminus I$ is union of open sets hence $I$ is closed.

1) and 2) are ok.