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It is said that $$\bigcup_{n\geq 1}\left(\frac 1n, 1+\frac1n\right)$$ is not compact.


Is it because it is not closed? Or am I missing something more?

Many thanks.

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@Hayden, shouldn't that be $0>n\geq1$ in that case? –  JMCF125 Apr 7 at 14:07
@JMCF125, I don't see how that inequality makes sense. Pre-edit the condition on $n$ was $n\in (0,1]$. Assuming that $n$ is an integer, then we'd only have $n=1$ as a possiblity, but then the union wouldn't be a cover of $(0,1]$, as many answers below use. On the other hand, if $n$ is not an integer, then $n\in (0,1]$ does make sense and does define a cover. Since everyone who answered treated $n$ as a positive integer, I had the question reflect that. –  Hayden Apr 7 at 14:13
@Hayden, sorry, I meant the opposite, $0<n\leq1$. The question said «for $(0,1]$» not «for $[1,+\infty)$». I assumed the OP meant "when $n$ is in... [that interval]" not "this equals... [the result]". –  JMCF125 Apr 7 at 14:22
I understand what it said, but I choose one notation as the reigning notation (i.e. that $n$ is natural) and made the condition on $n$ such that the resulting union was indeed a cover. If you feel that it would be better written as $\bigcup_{x\in (0,1]}{\left( \frac{1}{x},\frac{1}{x}+1\right)}$, then feel free to change it. Either one is fine, but as I said, the answerers predominantly treated $n$ as a positive integer, and thus wanted the question to reflect that. –  Hayden Apr 7 at 14:27

3 Answers 3

Here are four ways to see that $(0,1]$ is not compact.

  1. The open cover you gave for $(0,1]$ (namely $\{(1/n,1+1/n)\,:n\in\mathbb N\}$ does not have any finite subset which covers $(0,1]$ (in other words, does not have a finite subcover). I think this is the reason you were looking for, as user44441 said.
  2. A subset of $\mathbb R^n$ is compact if and only if it is closed and bounded. $(0,1]$ is not closed (although it is bounded).
  3. Expanding on LAcarguy's comment, in a metric space ($\mathbb R$ is a metric space) a subset is compact if and only if it is sequentially compact: every sequence of the subset has a convergent subsequence. The sequence $1,1/2,1/3,\dots$ is contained in $S$ but each of its subsequences converges to $0$ and $0\notin(0,1]$.
  4. If $(0,1]$ were compact, it would be true that every continuous function $f: (0,1]\to\mathbb R$ attains a maximum and a minimum. But the function $f(x)=1/x$ defined on $(0,1]$ is continuous and unbounded.
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+1 Nice answer. –  Felix Marin Mar 23 at 4:17
+1 from me too. –  J.W.S Apr 2 at 8:08

One way to see that (0, 1] is not compact is that 0 is a limit point of the set but it is not in the set.

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That specific union is probably meant to show you from the definition that it is not compact i.e., it is an open cover of $(0,1]$ which has no finite sub-cover. Because any finite sub-cover would have a lower bound $1/N$, for some $N$ and then this sub-cover would necessarily miss $(0, 1/N]$.

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There is some subtlety here about how $(0,1]$ is defined. If OP wants to say that it is not a compact subset of $\mathbb R$ that is one thing - and I think that is what may be in view, because the open sets which form the cover are not subsets of $(0,1]$. But a different way of looking at $(0,1]$ is that it inherits a topology from $\mathbb R$ of which it is a subset. The cover would then be sets of the form $(\frac 1n,1]$ because there would be nothing outside the interval $(0,1]$. –  Mark Bennet Mar 22 at 20:22

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