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Prove that $[0,1]$ is not a compact subset of $\mathbb{R}$ with the lower limit topology, i.e. open sets are of the form $[a,b)$.

My question is will different topology affect compactness of a set? If this is so, why? At first, when I see this question, I thought something is wrong with this question because I know that $[0,1]$ is compact by using Heine-Borel theorem.

Idonknow
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  • To give a trivial (literally) example to show how different topologies can affect compactness, let $X$ be any infinite set, and consider $\mathcal T_1$ the discrete topology on $X$ (all subsets of $X$ open) and $\mathcal T_2$ the indiscrete topology on $X$ (only $\emptyset$ and $X$ are open). Then $X$ is not compact under the topology $\mathcal T_1,$ but is compact under the topology $\mathcal T_2$. Roughly speaking, a topological space is compact if we don't have "too many" open sets, in a sense. – Cameron Buie Oct 22 '13 at 16:47
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    So if $\tau_1 \subset \tau_2$, and $X$ is not compact under the topology $\tau_1$, then $X$ is also not compact under $\tau_2$? – Idonknow Oct 22 '13 at 17:31
  • Yes, indeed! Well-spotted. – Cameron Buie Oct 22 '13 at 17:35
  • Is there such a theorem which say something similar to this? – Idonknow Oct 22 '13 at 17:36
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    Depends on what you mean by Theorem. If you just mean "provably true statement," then, yes. If you mean "major result," then that's debatable. If you mean "famous result that has a standard name associated with it," then no. It is a nice result, though, since every $\tau_1$-open cover is a $\tau_2$-open cover, so if there is a $\tau_1$-open cover without finite subcover, then there is a $\tau_2$-open cover without finite subcover. – Cameron Buie Oct 22 '13 at 17:40

4 Answers4

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In that case you can't use the Heine-Borel theorem because this theorem only apply to the case of normal topology.

To prove this proposition, you just find the open cover of $[0,1]$ such that every finite subcover does not cover $[0,1]$. Let $\mathcal{A}$ be a set of open sets defined as $$\mathcal{A}=\{[0,r):0<r<1\}\cup \{[1,2)\}$$ then $\mathcal{A}$ cover $[0,1]$. However every finite subcover of $\mathcal{A}$ does not cover $[0,1]$.

Hanul Jeon
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  • How do we know heine borel theorem is only applicable to standard topology? – Idonknow Oct 22 '13 at 17:22
  • @Idonknow In extreme case, consider the $\Bbb{R}$ with the discrete topology (i.e. every point is open set.) In discrete space, subspace $X$ is compact iff $X$ is finite. It is very different result for the Heine-Borel theorem. – Hanul Jeon Oct 22 '13 at 22:29
  • Suppose I have a square in $\mathbb{R^2}$ including the interior and it has subspace topology, can I use heine borel theorem to conclude that the square is compact since it is closed and bounded ? – Idonknow Oct 23 '13 at 04:23
  • @Idonknow 'Boundness' is depend on the metric of the given space. It is not depend on the topology. In that case, you have some caution to apply this theorem. For example, the set $(0,1)^2$ is closed bounded subspace of $(0,1)^2$ (if you give the subspace topology on $(0,1)^2$. It is closed because it is whole space.) but it is not compact. – Hanul Jeon Oct 23 '13 at 04:53
  • So in my question which involves square, I cannot apply heine borel theorem to conclude the square is compact? – Idonknow Oct 23 '13 at 05:02
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    @Idonknow If you give the normal subspace topology, you can use it but you need to caution to use it. If $A$ is a subspace of $\Bbb{R}^2$ and $B\subset A$, then $B$ is compact iff $B$ is a closed on $\Bbb{R}^2 and bounded. (not $A$! even if $B$ is closed in $A$ there is no reason that $B$ is closed in $\Bbb{R}^2$.) – Hanul Jeon Oct 23 '13 at 05:07
  • Why we need to make sure B is closed in the whole space? I very confused about closed set now. Because in heine borel theorem, it doesn't specify the set must be closed in where. – Idonknow Oct 23 '13 at 05:18
  • @Idonknow Boundness makes the problem. Boundness is not topological property. Do you know the proof of the Borel-Heine theorem? Borel-Heine theorem is consequence of 'every closed ball $\overline{B}(a,r){x:d(x,a)\le r}$ is compact, and if $A$ is a closed bounded (in $\Bbb{R}^2$!) subspace in $\Bbb{R}^2$ then $A\subset\overline{B}(a,r)$ for some $a$, $r$, so $A$ is a closed subspace of compact space. – Hanul Jeon Oct 23 '13 at 06:17
  • It implies compactness of $A$. But in the case of (generally not closed) subspace $X\subset\Bbb{R}^2$ and closed subset $Y\subset X$ in $X$, there is no reason that $Y$ is closed in $\overline{B}(a,r)$ for $a\in \Bbb{R}^2$ and $r>0$. – Hanul Jeon Oct 23 '13 at 06:18
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It is not compact : Take the open cover $$ \{(-\infty, 1-1/n)\}_{n=1}^{\infty} \cup \{[1,\infty)\} $$ In fact, compact sets in this topology are necessarily countable.

So yes, compactness certainly depends on the topology.

Prahlad Vaidyanathan
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Two answers have already demonstrated that the interval is not compact, by exhibiting simple open covers for it with no finite subcover.

The other half of the question is why compactness depends on the topology. I think in this case it might be instructive to consider the discrete topology. In the discrete topology, no infinite set $S$ is compact, because one can exhibit the open cover consisting of the sets $\{s\}$ for each $s\in S$. Since each point of $S$ is contained in exactly one of the elements of the cover, none of the elements can be omitted, and the cover not only has no finite subcover, it has no proper subcover at all! So in the discrete topology, a set is compact if and only if it is finite, which is quite different from the situation in a metric space.

The situation in the indiscrete topology is different in exactly the opposite way; every set is compact. The only nonempty set is the entire space, so there are no infinite open covers to begin with; every open cover consists of just this one open set, and a finite cover obviously has a finite subcover.

MJD
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GRE9367 #62

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Ian Coley's solution:

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Sean Sovine's solution:

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To prove $X$ is not compact, my first proof was similar to Ian Coley's, but I came up with another proof:

If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $\mathbb R$. This implies that the topologies are equal by (*), a contradiction.

(*) Munkres Exer26.1 (dbfin pf)

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BCLC
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