I'm not sure how to start this:

Is K := {$\frac{1}{n+1}$| n ∈ N} compact on $T_{up}$, $T_{k}$ and $T_{uplim}$?

I get that $T_{st}$ is not compact since it doesn't contain the limit point 0, and $T_{up} \subset T_{st} \subset T_{k} \subset T_{uplim}$, as the topology gets "finer", it's harder to admit finite sub-collection of coverings for K, is this correct? And how to put this in a formal way?

For $T_{up}$, I think it's not compact because it will never cover 0. Please give me some idea on how to do this one.

Definitions: $B_{st}$ stands for the space for standard topology

K topology := The collection $B_K$:=$B_{st}$ ∪ {(a, b) \ K}, where K := {1n∈N>0} defines a basis for a topology on R

Upper topology:= {∅, R}∪{(a, ∞) | a ∈ R} defines a topology on R

Upper limit topology:= {(a, b]} defines a basis for a topology on R.

Thank you so much for the help!

  • $\begingroup$ $\mathcal{T}_{\text{up}}$ is a subset of $\mathcal{T}_{\text{st}}$, not the other way round. $\endgroup$ – Henno Brandsma Mar 1 '18 at 8:54
  • $\begingroup$ @HennoBrandsma I wrote $T_{st}$ is a subset of the upper limit topology, but I will update my question to include $T_{up}$ $\endgroup$ – Liz Mar 1 '18 at 13:50

It's good to always have the exact definition of compactness in mind when you are working with problems of this sort.

A topological space $X$ is compact if every open cover of $X$ contains a finite subcover.

Conversely, to show $X$ is not compact, it suffices to find an open cover that does not admit a finite subcover.

$K$ is not compact in the standard topology on $\mathbb{R}$ precisely because we can find an open cover of $K$ that has no finite subcover. If $$K = \left\{1, \frac12, \frac13, \frac14, \ldots \right\},$$ then the collection of open intervals $$\left\{\left(1, 2\right), \left(\frac12, 2 \right), \left(\frac13, 2\right), \left(\frac14, 2 \right), \ldots \right\}$$ is an open cover of $K$ since $\frac1n \in \left(\frac{1}{n+1}, 2\right)$.

Suppose a finite subcover existed, it would have the form $$\left\{\left(\frac{1}{n_1}, 2\right), \ldots, \left(\frac{1}{n_m}, 2\right) \right\},$$

where I carefully ordered the intervals so that $n_1 < n_2 < \ldots < n_m$. Unfortunately, the point $\frac{1}{n_m + 1}$ is not covered, so $K$ is not compact.

Can you construct a similar open cover of $K$ in the upper topology on $\mathbb{R}$, so that it's easy to show that certain points are not covered by a finite subcover?

  • $\begingroup$ Thank you! Your proof makes a lot of sense to me $\endgroup$ – Liz Mar 1 '18 at 21:44
  • $\begingroup$ @Liz Glad I could help! $\endgroup$ – Andrey Portnoy Mar 2 '18 at 4:01

A formal way to see that $K$ is not compact in $T_{K}$ ($\subset T_{uplim}$) is, just as you said, by contradiction, if you supose $K$ is compact then every open cover for $K$ has a finite subcover, specially for covers formed by elements of $T_{st}\subset T_{K}$ ($\subset T_{uplim}$).

To see that $K$ is not compact with $T_{up}$ prove that the open cover $\{(\frac{1}{n},\infty)\}_{n\in\mathbb{N}^{+}}$ has not open subcovers.


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