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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.

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  • $\begingroup$ Stylistically the use of $0.5$ would be better written as $\frac{1}{2}$ since decimal notation is atypical in proof based mathematics. $\endgroup$ May 10 '18 at 20:09
  • $\begingroup$ @CyclotomicField, Useless quibble comparied to the glaring mistake of 1). $\endgroup$ May 10 '18 at 20:15
  • $\begingroup$ @WilliamElliot, can you please point out the apparent mistake made in 1)? Thank you. $\endgroup$
    – Alea
    May 10 '18 at 20:43
  • $\begingroup$ @A.Asad. A bracket is misplacec. $\endgroup$ May 11 '18 at 1:36
  • $\begingroup$ The right notation for the set in 1) is $G=\{x\in \Bbb R| 0<x<1\}.$ Computerese is not mathematical notation. For 3) state that for any $e>0$ the set $(-e,e)=V_e(0) $ is not a subset of $[0,1]$ because $ (-e,0)$ is a non-empty subset of $V_e(0)$ containing only negative numbers, which don't belong to $[0,1].$ For 4) use a similar method as in 1); For $0\ne x\in \Bbb R,$ consider the set $\{y\in \Bbb R: |y-x|<|x|\}. $ Note that I used a colon ($:$) instead of a vertical line ($|$), which I prefer; otherwise the other vertical lines (for the absolute values) make it hard to read. $\endgroup$ May 11 '18 at 2:38
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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.

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  • $\begingroup$ How would I prove 1)? Also, for 3), do you mean that $\forall \epsilon>0$, the epsilon neighbourhood of 0 would never be a subset of $[0,1]$? $\endgroup$
    – Alea
    May 10 '18 at 20:51
  • $\begingroup$ 1) Take $y$ s.t. $x-\varepsilon < y <x+\varepsilon$ and prove that $0<y<1$ using the value you chose for $\varepsilon$. 3) Yes, that's what I mean. $\endgroup$
    – paf
    May 10 '18 at 21:20
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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) $.

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  • $\begingroup$ Thank you! This is through another result. I was hoping to go through definitions for proofs. $\endgroup$
    – Alea
    May 10 '18 at 20:46
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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.

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  • $\begingroup$ What middle point, for 3? It's not open because of the endpoints, no? $\endgroup$
    – Alea
    May 10 '18 at 20:45
  • $\begingroup$ Suppose an open interval (a, b) contains 0 and is subset of [0, 1], that can't be since (a, b) contains points to the left of 0, then think of the middle point between a and 0 and inequalities $\endgroup$
    – id500
    May 10 '18 at 23:22

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