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$\newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\T}{\mathcal{T}} \newcommand{\R}{\mathbb{R}}$ Problem

Let $\T_1$ be the lower limit topology in $\R$ and $\T_{2}$ be the topology generated by basis $\mathcal{C}=\{[a,b)|a<b,a,b\in\Q\}$. Determine the closures of the intervals $A=(0,\sqrt{2})$ and $B=(\sqrt{2},3)$ in these two topologies.

Attempted Solution

（I shall use $A'$ to denote the set of limit points of set $A$）

In $\T_{1}$, we claim that $\overline{\left(0,\sqrt{2}\right)}=[0,\sqrt{2})$. To show this, we first show that $\left\{ 0\right\} $ is a limit point of $\left(0,\sqrt{2}\right).$ Let $U$ be any open set containing in $\left\{ 0\right\} $, then $\exists[a,b)$ such that $\left\{ 0\right\} \in[a,b)\subset U.$ Therefore, $\exists\varepsilon>0$ such that $[0,\varepsilon)\in[a,b),$ which implies $U\cap\left(0,\sqrt{2}\right)\neq\emptyset$. Since $U$ is arbitrary, we have that $\left\{ 0\right\} \in\left(0,2\right)'$. Then we have $[0,\sqrt{2})\subset\overline{\left(0,\sqrt{2}\right)}.$ To see the other direction, it suffices to show that $[0,\sqrt{2})$ is closed. Indeed, $[0,\sqrt{2})^{c}=\left(-\infty,0\right)\cup[\sqrt{2},+\infty)=\cup_{n=1}^{\infty}\left[[-n,0)\cup[\sqrt{2},n)\right]$ which is open and thus $[0,\sqrt{2})$ is closed and contains $\left(0,\sqrt{2}\right)$. Then by definition, $\overline{\left(0,\sqrt{2}\right)}\subset[0,\sqrt{2})$. By virtually the same argument, we can show that $\overline{\left(\sqrt{2},3\right)}=[\sqrt{2},3)$.

In $\T_{2}$, we claim that $\overline{\left(0,\sqrt{2}\right)}=\left[0,\sqrt{2}\right].$ Using a similar argument, we can show that $\left\{ 0\right\} \in\left(0,\sqrt{2}\right)'$. To show that $\left\{ \sqrt{2}\right\} \in\left(0,\sqrt{2}\right)'$ , we note that for any open set $U$ containing $\left\{ \sqrt{2}\right\} $, $\exists[a,b)\in U$ such that $a<b$ and $a,b\in\Q.$ This guarantees that $a<\sqrt{2}$ since $\sqrt{2}\notin\Q$. Thus $U\cap\left(0,\sqrt{2}\right)\neq\emptyset.$ Since $U$ is arbitrary, $\sqrt{2}\in\left(0,\sqrt{2}\right)'$. We then have $\left[0,\sqrt{2}\right]\in\overline{\left(0,\sqrt{2}\right)}.$ To see this other direction, it suffices to show that $\left[0,\sqrt{2}\right]$ is closed. To see this, we note that we can rewrite $[0,\sqrt{2}]^{c}=\cup_{n=1}^{\infty}\left[[-n,0)\cup[a_{n},+\infty)\right]$ where $a_{n}\in\Q$ for all $n\in\N,$ and $a_{n}\downarrow\sqrt{2}.$ Notice that the latter is open as it's a countable union of open sets in $\T_{2}$, whence $[0,\sqrt{2}]$ is closed, and as a result $\overline{\left(0,\sqrt{2}\right)}\subset[0,\sqrt{2}]$. By a similar argument, we can show that $\overline{\left(\sqrt{2},3\right)}=[\sqrt{2},3)$

Question

I don't know if my arguments are convincing enough. I hope someone can take a look at my attempt and point out any mistakes and possible improvements.

Martin Sleziak
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Mathemagician
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  • I would give the details showing that $\sqrt2$ is in the closure of $(\sqrt2,3)$ in the second topology, since the argument is just a little different, but otherwise it's quite clear. – Brian M. Scott Oct 28 '15 at 04:09
  • @BrianM.Scott Thanks for pointing that out. I think a similar argument can be applied to show that $\sqrt{2}$ is a limit point because any rational interval of the form $[a,b)$ must contain $[\sqrt{2},\epsilon)$ for some $\epsilon >0$. The fact that ${3}$ isn't a limit point is straightforward from $[3,+\infty)\cap (\sqrt{2},3)=\emptyset$. We can show that $[\sqrt{2},3)$ is closed by taking ${a_n}\in\Q$ such that $a_n\uparrow\sqrt{2}$ – Mathemagician Oct 28 '15 at 04:28
  • Yes your argument is absolutely correct –  Oct 28 '15 at 04:30
  • @Rememberme Thanks! – Mathemagician Oct 28 '15 at 04:32
  • One little refinement : You can drop the word "countable" in "it's a countable union of open sets". The union of any family of open sets is open. – DanielWainfleet Oct 28 '15 at 19:14

1 Answers1

1

$\newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\T}{\mathcal{T}} \newcommand{\R}{\mathbb{R}}$The definitions of $\mathcal{T}_1$ and $\mathcal{T}_2$ are similar. The generated by intervals of the real line:

  • $\mathcal{T}_1$ generated by half-open intervals $\{[a,b)|\,a<b \text{ with }a,b\in\R\}$
  • $\mathcal{T}_2$ generated by half-open intervals $\{[a,b)|\,a<b \text{ with }a,b\in\Q\}$

So what difference does it make if we only use rational numbers $\mathbb{Q}$ instead of every possible real number $\mathbb{R}$?

In this system, can the closures contain the left points? $0 \in \overline{(0, \sqrt{2})}$ or $\sqrt{2} \in \overline{(\sqrt{2}, 3)}$ ?

  • if $0 \in [a,b)$ then $ a \leq 0 < b$, so that $[a,b)$ has some positive numbers

  • if $\sqrt{2} \in [a,b)$ then $ a \leq \sqrt{2} < b$ so the interval $[a,b)$ can't avoid sharing some points in $(\sqrt{2},3)$

Both cases had the same outcome, regardless if we used $\mathbb{R}$ or $\mathbb{Q}$... What about the points on the right?

  • if $\sqrt{2} \in [a,b)$ then $ a \leq \sqrt{2} < b$ but if $a \in \mathbb{Q}$ then $a < \sqrt{2}$ since $\sqrt{2}$ is not a fraction.

    • $(0, \sqrt{2}) \cap [\sqrt{2}, \infty) = \varnothing $
    • This is acceptable in $\mathcal{T}_1$ but not $\mathcal{T}_2$.

Let's draw the perfect non-overlap in this case, that both $\mathcal{T}_1$ always permits but $\mathcal{T}_2$ only sometimes:

[oooooooooooooooooooo)................
.....................[ooooooooooooooo)

vs                   3?

(oooooooooooooooooooo)................
.....................(ooooooooooooooo)
Martin Sleziak
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cactus314
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