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?

Thanks in advance!

  • 1
    Your definition of $\tau$ doesn't make sense: you are using $\tau$ to define it! – José Carlos Santos Oct 11 at 11:13
  • “There exists $x, y$ such that some property holds, but also does not hold?” – Joppy Oct 11 at 11:14
  • 1
    each point of your set if isolated, so you get the set itself, and the only limiting point which is $0$. – Hayk Oct 11 at 11:17
  • 1
    @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 at 11:25
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    @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 at 11:29
up vote 1 down vote accepted

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 $\frac{1}{m}<a$. 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 $0\in\overline{\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}}\neq\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\frac{1}{n},...\}$

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