Let $S=(1,2)$; describe an open cover of S that has no finite subcover.


If $E$ is a set, and {$G$$_\alpha$}$_{\alpha\in A}$ is a collection of sets such that $E \subset \bigcup_{\alpha\in A}$$G$$_\alpha$, then the collection {$G$$_\alpha$}$_{\alpha\in A}$ is called a cover of $E$.

If {$G$$_\alpha$}$_{\alpha\in A}$ is a cover of $E$ and $B\subset A$ such that $E \subset \bigcup_{\alpha\in B}$$G$$_\alpha$ then the collection {$G$$_\alpha$}$_{\alpha\in B}$ is called a subcover of $E$.


$\{\;(1,2-2^{-n})\;:n\in \mathbb N\}.$

  • $\begingroup$ Can you please elaborate on how this will satisfy being an open cover that has no finite subcover? $\endgroup$ – user2553807 Feb 19 '17 at 16:19
  • $\begingroup$ If $D$ is a non-empty finite subset of it, and $x=\max \{y: (1,y)\in D\}$ then $\cup D=(1,y)\ne (1,2).$ $\endgroup$ – DanielWainfleet Feb 19 '17 at 18:04


$\{(1.5 - \epsilon, 1.5 + \epsilon)|0< \epsilon < .5\}$. That's an uncountable cover (that does have a countable subcover; e.g. restricting $\epsilon$ to rationals).


$\{(1+\frac{1}{k+1}, 1 + \frac 1{k-1})|k \in \mathbb N; k \ge 2\} = \{(1\frac 13, 2),(1\frac 14, 1\frac 12),(1\frac 15,1 \frac 13),....\}$

or another $\{(r - \epsilon, r + \epsilon)| 1 < r < 2; \epsilon =\frac{\min(r-1, 2 -r)}2 = \text {half the distance from r to the nearest endpoint of (1,2)}\}$

In both of these, if there were a finite subcover, there would be a smallest (yet positive) $\epsilon$ or an largest (yet finite) $k$ so that $(1, 1.5-\epsilon]\cup [1.5+ \epsilon)$ or $(1, \frac 1{k+1}]$ are not covered.

(Similarly in the third, a finite subcover is there'd be a lowest and highest $r$ so there will always be a gap between the lowest or highest point covered and the "edge". That should give you an intuition that, in general, that a compact set must be closed. For a limit point not in the set there are an infinite number of open sets that can get close to the limit point without including it without there being a closest such set.)

(So to come up with a counter example, just think of a sequence that converges to the point and the the open cover is all the open sets "beyond" the point. Ex. $\{(1 + \epsilon, \infty)\}$ or $\{(-\infty,2 -\epsilon)\}$ etc.)


The first the subcovers are all "nested" so that the you can remove any of the "early ones". Intuitively only the "last" one matters but because there is no last one there can not be a finite subcover.

The second one is an interlocked chain of increasingly small intervals. If you remove any one though you no longer have a cover as each interval contains one point (the upper and lower endpoint of the other two intervals "next to" it) that is not contained in any other.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.