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Consider the open cover $\mathscr{F}=\begin{Bmatrix}\pmatrix{0,5-\frac{1}{n}}:n\in\mathbb{N}\end{Bmatrix}$ for the set $(0,5)$.

Question:

Is $(0,5)=\bigcup^\infty_{n=1}\pmatrix{0,5-\frac{1}{n}}$? Why does $\mathscr{F}$ have no finite subcover?

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  • $\begingroup$ It would seem that if you take any interval out of the infinite union you'd still cover $(0,5)$. $\endgroup$ Sep 12, 2013 at 4:18
  • $\begingroup$ But what happens if you only consider finitely many? $\endgroup$ Sep 12, 2013 at 4:19
  • $\begingroup$ OK, so since no finite subset of $\mathscr{F}$ can possibly cover the infinite set $(0,5)$, we say that "$\mathscr{F}$ has no finite subcover over $(0,5)$," right? $\endgroup$ Sep 12, 2013 at 4:22
  • $\begingroup$ @GrumpyParsnip, do you know what the TAPI pipeline is? $\endgroup$ Sep 12, 2013 at 4:23
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    $\begingroup$ Suppose $\mathscr{G}$ were a finite subcover. Let $N\in\mathbb{N}$ be the largest $N$ such that $\pmatrix{0,5-\frac{1}{N}}$ is in $\mathscr{G}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, choose a rational $q\in\mathbb{Q}$ such that $5-\frac{1}{N}<q<5$. This contradicts that $\mathscr{G}$ covers $(0,5)$. $\endgroup$ Sep 12, 2013 at 4:28

1 Answer 1

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$\mathscr{F}$ is an open cover of $(0,5)$; it is not a cover of $\{(0,5)\}$.

  1. Yes, $(0,5)=\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$. On the one hand it’s clear that $\left(0,5-\frac1n\right)\subseteq(0,5)$ for each $n\in\Bbb Z^+$, so $\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)\subseteq(0,5)$. On the other hand, if $x\in(0,5)$, then $x<5$, and $5-x>0$. Thus, there is a positive integer $k$ such that $\frac1k<5-x$ and hence $x<5-\frac1k$, so that $x\in\left(0,5-\frac1n\right)\subseteq\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$. Since $x$ was an arbitrary element of $(0,5)$, it follows that $(0,5)\subseteq\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$ and hence that $(0,5)=\bigcup_{n=1}^\infty\left(0,5-\frac1n\right)$.

  2. Let $\mathscr{F}_0$ be any finite subset of $\mathscr{F}$, say $\mathscr{F}_0=\left\{\left(0,5-\frac1{n_k}\right):k=1,\ldots,m\right\}$. Let $n=\max\{n_1,\ldots,n_m\}$; then $5-\frac1{n_k}\le\frac1n$ for $k=1,\ldots,m$, so $$\left(0,5-\frac1{n_k}\right)\subseteq\left(0,5-\frac1n\right)$$ for $k=1,\ldots,m$. This implies that $$\bigcup_{k=1}^m\left(0,5-\frac1{n_k}\right)=\left(0,5-\frac1n\right)\subsetneqq(0,5)$$ and hence that $\mathscr{F}_0$ does not cover $(0,5)$. That is, no finite subset of $\mathscr{F}$ covers $(0,5)$, i.e., $\mathscr{F}$ is an open cover of $(0,5)$ with no finite subcover.

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  • $\begingroup$ The professor here use $\{(0,5)\}$ to mean $(0,5)$. $\endgroup$ Sep 12, 2013 at 4:29
  • $\begingroup$ See my $\mathscr{G}$-argument in the comments below my question. $\endgroup$ Sep 12, 2013 at 4:32
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    $\begingroup$ @SecureSpace: I doubt very much that any Harvey Mudd professor makes such an elementary mistake. That lecture is more than an hour long; if you give me a time stamp for a point at which you think that he’s made that mistake, I’ll take a look. In the meantime, I strongly suspect that he’s used $\{(0,5)\}$ correctly, to mean the set whose only member is the open interval $(0,5)$. This is of course a finite open cover of $(0,5)$, though not a very interesting one. $\endgroup$ Sep 12, 2013 at 4:35
  • $\begingroup$ @SecureSpace: Yes, the argument with $\mathscr{G}$ works fine, though there’s no need to pick a rational number: any number in $\left[5-\frac1N,5\right)$ will do fine. In fact, you can simply use $5-\frac1N$. $\endgroup$ Sep 12, 2013 at 4:36
  • $\begingroup$ 00:15:14, let me know if you're still in doubt. $\endgroup$ Sep 12, 2013 at 5:04

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