Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I have shown that the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.

To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?

For example, we have that $\displaystyle{\tilde{S}_1=\left \{\left (\frac{x}{4}, \frac{3x}{4}\right ): 0<x<1\right \}}$ is a subset of $S_1$, isn't it?

This is not a cover of $A$, since for $n=1$ the element $a$ is not in $\tilde{S}_1$. The same holds also when we consider a subset of the form $\left \{\left (\frac{x}{i}, \frac{3x}{i}\right ): 0<x<1\right \}$ with $i>2$.

Does this mean that $S_1$ has no finite subcover?

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    $\begingroup$ You are right that all these families are subsets of $S_1$, but this does not mean that $S_1$ has no finite subcover. But you indeed cannot find a finite subcover in $S_1$, and to see this you should note that otherwise some of $(x/2,3x/2)$ will contain all $1/n$ starting from some $n_0$. Now you can obtain a contradiction easily. $\endgroup$ – SMM Apr 28 '18 at 16:57
  • $\begingroup$ Do you mean that the set $A$ is not finite, and that's why we cannot find a finite subcover? @SMM $\endgroup$ – Mary Star Apr 28 '18 at 17:26
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    $\begingroup$ No. For any finite subcover, say consisted of $(x_i/2,3x_i/2)$ for $0\leq i\leq k$, you can find $n$, such that $1/n$ is less than all $x_i/2$. This is because all $x_i>0$, but $\lim_n 1/n=0$. So, this subcover does not covers $1/n$. Since the subcover was arbitrarily chosen, this means that finite subcover does not exist. (This is not what I first had in mind, but I now see that it is more clear.) $\endgroup$ – SMM Apr 28 '18 at 17:32
  • $\begingroup$ What is "the element $a$"? $\endgroup$ – DanielWainfleet Apr 28 '18 at 18:03
  • $\begingroup$ Oh, it is $a=\frac{1}{n}$ for some $n$. @DanielWainfleet $\endgroup$ – Mary Star Apr 28 '18 at 18:11

Let $F$ be a non-empty finite subset of $S_1$. Then $F=\{(x/2,3x/2): x\in G\}$ where $G$ is a finite non-empty subset of $(0,1).$ Let $y=\min \{x/2:x\in G\}.$

Then $y>0.$ If $z$ is any member of $ \cup F$ then for some $x\in G$ we have $z\in (x/2,3x/2)$ so $z>x/2\geq y.$

Take $n\in \Bbb N$ with $1/n<y.$ Then $1/n$ is less than every member of $ \cup F.$ So $1/n\not \in \cup F$. So $F$ is not a cover of $A. $

  • $\begingroup$ On the other hand, if $x=1/2$ then $(x-1/2,x+1/2)=(0,1)\in S_2.$ And $A\subset (0,1).$ So $\{(0,1)\}$ is a $1$-member subset of $ S_2$ and is a cover of $A.$ $\endgroup$ – DanielWainfleet Apr 28 '18 at 18:21
  • $\begingroup$ Why do we take $\frac{1}{n}<y$ ? $\endgroup$ – Mary Star Apr 28 '18 at 18:58
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    $\begingroup$ We have $\cup F\subset (y,\infty)$ so $\cup F$ is disjoint from $(0,y).$ But $1/n\in (0,y)\cap A$ if $n\in \Bbb N$ is large enough. So $A$ has members that don't belong to $\cup F$..... So $A$ is not a subset of $\cup F$.... Therefore $F$ is not a cover of $ A.$....This applies to $any$ non-empty finite $F\subset S_1$. $\endgroup$ – DanielWainfleet Apr 28 '18 at 19:17
  • $\begingroup$ How can we justify formally that $\frac{1}{n}\in (0,y)$ for large $n$ ? $\endgroup$ – Mary Star Apr 28 '18 at 21:06
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    $\begingroup$ Yes we can say that. $\endgroup$ – DanielWainfleet Apr 29 '18 at 3:32

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