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.

  • $\begingroup$ @Hayden, shouldn't that be $0>n\geq1$ in that case? $\endgroup$ – JMCF125 Apr 7 '14 at 14:07
  • $\begingroup$ @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. $\endgroup$ – Hayden Apr 7 '14 at 14:13
  • $\begingroup$ @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]". $\endgroup$ – JMCF125 Apr 7 '14 at 14:22
  • $\begingroup$ 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. $\endgroup$ – Hayden Apr 7 '14 at 14:27
  • $\begingroup$ Nice reading for beginners: blogs.scientificamerican.com/roots-of-unity/…. $\endgroup$ – tatan Sep 21 '20 at 18:09

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.

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.


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 $(0, 1/N]$.

  • 1
    $\begingroup$ 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 $(0,1]$. $\endgroup$ – Mark Bennet Mar 22 '14 at 20:22

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