0

I am trying to show the following set inclusion holds for $X:=\mathbb{R}$,

$\mathscr{A}:=\left\{A\subseteq X:\exists U\in\tau\text{ such that } 0\in U\subseteq A\right\} \subseteq \left\{A\subseteq X:X\setminus A\text{ is finite}\right\}=:\mathscr{A}',$

where the basic nhbds at the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$.

This is something I need to hold true, as part of a larger problem. My attempt is as follows,

Let $A\in \mathscr{A}$.

By definition of the set,

$$\exists U\in\tau \text{ such that } 0\in U\subseteq A.$$

By definition of the basic nhbds at the origin,

$$U:=\left(\bigcup_{\epsilon>0}(-\epsilon, \epsilon)\right) \cup (-\infty, -n) \cup (n, \infty).$$

Therefore, $X\setminus A\subseteq X\setminus U$, and by De Morgan,

\begin{align*} X\setminus U&=\left(\bigcup_{\epsilon>0}(-\epsilon, \epsilon)\right)^c \cap (-\infty, -n)^c \cap (n,\infty)^c. \\ \end{align*}

This is where I get stuck. I see that if I can show that $X\setminus U$ is finite, then I can conclude that $X\setminus A$ is finite and thus $A\in \mathscr{A}'$ giving the desired inclusion. I have a feeling I may be taking $U$ to be of the wrong form however.

1 Answers1

2

The claim is not true: for instance $(-1/2,1/2)\cup(-\infty,-1)\cup(1,\infty)\in\mathscr{A}\setminus\mathscr{A}^\prime$.

  • But that set does not belong to $\mathscr{A}$? Because you chose $\epsilon=1/2$, but it must hold for any choice of $\epsilon.$ That is, it must be that, for instance, $(-3/4,3/4)\cup(-\infty,-1)\cup(1,\infty)\subseteq(-1/2,1/2)\cup(-\infty,-1)\cup(1,\infty)$. Maybe I am missing something, or confused. –  Feb 02 '21 at 15:42
  • A neighbourhood of $0$ is just a set $N$ such that there is some $V\in\tau$ such that $0\in V\subseteq N$. The fact that the family $\mathfrak F:={(-\varepsilon,\varepsilon)\cup(-\infty,-n)\cup (n,\infty),:, \varepsilon>0,\ n\in \Bbb N}$ is a family of basic neighbourhoods of $0$ just means every $N\in\mathfrak F$ is a neighbourhood of $0$, and that for all $V\in\tau$ either $0\notin V$ or there is some $M\in\mathfrak F$ such that $M\subseteq V$. Regardless of the latter, you can see by the first fact I've mentioned that every neighbourhood of $0$ is in $\mathscr A$. –  Feb 02 '21 at 15:48
  • Okay thanks. This question is really apart of a bigger question, which I guess I am interrupting wrong. The larger question is really-- Trying to show that $X$ the real line, with this topology as defined, then the Fréchet filter converges to $0$. Could you help interrupt what I need to show? –  Feb 02 '21 at 15:52
  • I am unfamiliar with filters. –  Feb 02 '21 at 15:53
  • No problem, I actually got mixed up with definition from my textbook and online notes. –  Feb 02 '21 at 16:09
  • @D.Math: The Fréchet (or cofinite) filter on $\Bbb R$ does not converge to $0$ in the topology in question. The Fréchet filter on $\Bbb N$ or $\Bbb Z$, on the other hand, does converge to $0$: every open nbhd of $0$ contains a cofinite subset of $\Bbb Z$ (and hence of $\Bbb N$). – Brian M. Scott Feb 03 '21 at 03:01