Closure of $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},…\frac{1}{n},…\}$

Exercise: What is the closure in $$\mathbb{R}$$of each of the following set:

i) $$\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$$

My attempt: I consider the topological space $$\mathbb{R},\tau$$, where $$\tau$$ is the following topology $$\tau=\mathbb{R}\cup\emptyset\cup\{(a,b):\forall a,b\in\mathbb{R}\}$$

The open sets can be defined as $$\{A\in\tau:\text{such that} \forall a \in A\:\exists x,y\in\mathbb{R}, a\in(x,y)\subseteq A\}$$

$$\forall i\in \mathbb{N},\frac{1}{i}\in\mathbb{R}$$

For elements $$\frac{1}{i_1}$$ and $$\frac{1}{i_1}$$ there exists $$x,y\in\mathbb{R}$$ such that $$\frac{1}{i_1}\in(x,y)$$ but $$\frac{1}{i_1}\notin(x,y)$$

Then $$\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$$ has not limit points then $$\overline{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}=\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$$

Question:

I am not really sure about the step "there exists $$x,y\in\mathbb{R}$$ such that $$\frac{1}{i_1}\in(x,y)$$ but $$\frac{1}{i_1}\notin(x,y)$$"

Is my proof right? If not is there any way to prove this?

• Your definition of $\tau$ doesn't make sense: you are using $\tau$ to define it! – José Carlos Santos Oct 11 '18 at 11:13
• “There exists $x, y$ such that some property holds, but also does not hold?” – Joppy Oct 11 '18 at 11:14
• each point of your set if isolated, so you get the set itself, and the only limiting point which is $0$. – Hayk Oct 11 '18 at 11:17
• @PedroGomes, "I can find an open interval that contains 0 but not 1" is not relevant to $0$ being a limiting point. You need to to show that any (small) neighborhood of $0$ contains a point from your set other than $0$. This is obviously true, making $0$ into the closure. – Hayk Oct 11 '18 at 11:25
• @PedroGomes Your new $\tau$ isn't a topology, since $(-1,0),(0,1)\in\tau$, but $(-1,0)\cup(0,1)\notin\tau$. – José Carlos Santos Oct 11 '18 at 11:29
As I understood your definition of $$\tau$$ is standart topology on $$\mathbb{R}$$. And you may say the base of it must be $$\{(a,b):\forall a,b\in\mathbb{R}\}$$. And now say any $$a\in \mathbb{R^+}$$. By using precible of Archimedes, you can find $$m>0$$ integer such that $$\frac{1}{m}. So $$\frac{1}{m}\in{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}$$ and $$\frac{1}{m}\in (-a,a)$$. Since $$(-a,a)$$ is an arbitrary neighbourhood of $$0$$ then $$0\in\overline{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}\neq\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$$