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Consider the following topologies on $\mathbb{R}$

$\mathcal{T}_1=$ the standard topology

$\mathcal{T}_2=$ the lower limit topology

$\mathcal{T}_3=$ the topology having as basis all open rays $(- \infty, a)$

$\mathcal{T}_4=$ the topology having as basis all open intervals $(a,b)$ and all one-point sets $\{c\}$, such that $c \in \mathbb{Q}$

Determine, for each of the topologies, which of the others it contains.

This is what I found (and please tell me if it's correct):

$\mathcal{T}_1 \subset \mathcal{T}_2 $

$\mathcal{T}_1 \supset \mathcal{T}_3 $

$\mathcal{T}_1 \subset \mathcal{T}_4 $

$\mathcal{T}_2 \supset \mathcal{T}_3 $

$\mathcal{T}_2$ and $\mathcal{T}_4 $ are not comparable

$\mathcal{T}_3 \subset \mathcal{T}_4 $

Let $A=(2,\sqrt{7})$. I need to determine the closure of $A$ in each of the above topologies.

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  • $\begingroup$ @AmitaiYuval: Really? So you say $\tau_1\subseteq \tau_3$? $\endgroup$ – tomasz Sep 18 '14 at 20:09
  • $\begingroup$ @AmitaiYuval: But that's what OP says, while you say you disagree. Unless you mean something else? $\endgroup$ – tomasz Sep 18 '14 at 20:20
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It is not so hard to see that $\tau_3\subset\tau_1\subset\tau_2$, and also $\tau_1\subset\tau_4.$ We now show that $\tau_2,\tau_4$ are incomparable.

Every non-empty basic open set in $\tau_2$ is infinite, thus no singleton can be open in $\tau_2$ and $\tau_4\not\subset\tau_2$.

On the other hand, consider the set $U=[\sqrt{2},11)$, which is open in $\tau_2$. The point $\sqrt{2}\in U$ does not have any basic open neighborhood of $\tau_4$, thus $U$ is not open in $\tau_4$ and $\tau_2\not\subset\tau_4$.

Here are a few facts that may ease finding the closure of $A=(2,\sqrt{7})$ in each one of the topologies:

For $\tau_1$ we know the answer.

Note that every half open interval $[a,b)$, besides being open in $\tau_2$, is also closed in $\tau_2$ (why?).

All non-trivial open subsets in $\tau_3$ are of the form $(-\infty,a)$, thus all (non_trivial) closed subsets are $[a,\infty)$.

The subset $(2,\sqrt{7}]$ is closed in $\tau_4$.

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  • $\begingroup$ Could you help with $\tau_3$ and $\tau_4$ $\endgroup$ – Carpediem Sep 19 '14 at 17:35
  • $\begingroup$ As written in my answer, the closed subsets in $\tau_3$ are $[a,\infty)$. You need the smallest of those, which contains $A$. $\endgroup$ – Amitai Yuval Sep 19 '14 at 19:29
  • $\begingroup$ So $\tau_3= (2,\infty )$ ? and $\tau_4= (2,\sqrt{7})$ $\endgroup$ – Carpediem Sep 20 '14 at 22:09
  • $\begingroup$ These subsets are not closed... $\endgroup$ – Amitai Yuval Sep 21 '14 at 5:49
  • $\begingroup$ So in $\tau_4$ it is $(2,\sqrt{7}]$. However in $\tau_3$ I still don't understand $\endgroup$ – Carpediem Sep 21 '14 at 18:52

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