Show that a subset of $\mathbb{R}$ is compact in upper limit topology

Question

I want to show that $A = [0,1]$ is not a compact subspace of $\mathbb{R}$, where $\mathbb{R}$ has the upper limit topology with open sets of the form $(a,b] = \{x \in \mathbb{R}\space|\space a < x \leq b\}$.

I read a suggested solution which confused me. It said that with the open cover $\mathcal{A} = \{(r, 1] \space | \space 0 < r \leq 1 \} \cup (-1,1]$ of $A$, there is a finite subcover that does not cover $A$. I understand that this is the case, but not why that would prove that $[0,1]$ is not compact. As far as I know every open covering should have a finite subcollection covering $A$. Not that every finite subcollection must cover $A$. If someone would like to clarify this, I would be very greatful.

Some suggestions on how to prove the statement that $A$ is not compact under this topology would also be helpful. I find compactness to be a confusing subject, to say the least...

Answer 1

I don't understand that remark. But you can prove that $(0,1]$ is not compact using the open cover$$(-1,0]\cup\left(\frac12,1\right]\cup\left(\frac13,\frac12\right]\cup\left(\frac14,\frac13\right]\cup\cdots$$Since it is a disjoint union, there no subcovers other than the original one.

Answer 2

For a topological space $X$ with a topology $\tau$ a subset $A\subset X$ is called compact in $X$ if and only if each open cover $\mathcal{A}\subset \tau$ of $A$ has a finite cover $\mathcal{B}\subset\mathcal{A}$ of $A$.

The solution you get is wrong and doesn't prove that $A=[0,1]$ is not compact in the upper limit topology of $\mathbb{R}$. Either the example of José do it or you can fix your solution if you consider $$ \mathcal{A}=\{(r,1]~:~0<r<1\}\cup(-1,0]. $$ You get $\bigcup_{I\in\mathcal{A}}I=(-1,1]\supset A$. But for each finite subset of $\mathcal{B}\subset\mathcal{A}$ you get either $$ \bigcup_{I\in\mathcal{B}}I=(-1,0]\cup(\tilde{r},1]\not\supset[0,1]=A $$ or $$ \bigcup_{I\in\mathcal{B}}I=(\tilde{r},1]\not\supset[0,1]=A $$ where $\tilde{r}=\min\{r\in(0,1)~:~(r,1]\in\mathcal{B}\}>0$ is well defined since $\mathcal{B}$ is finite. This proves that $A=[0,1]$ is not compact in the upper limit topology.

Answer 3

This is similar to the lower limit topology version:

GRE9367 #62

Ian Coley's solution:

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 (1), a contradiction.

Here we have:

If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and upper limit topologies of $\mathbb R$. This implies that the topologies are equal by (1) and (2), a contradiction.

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(1) Munkres Exer26.1 (dbfin pf)

(2) Which one is finer: standard topology or upper limit topology? Both upper limit and lower limit topologies are finer than standard topology.