0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $\mathcal{T}_{\text{up}}$ 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
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

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.