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Here is the question I have:Consider the topologies $T_{disc}, T_{codisc}, T_{fin}, T_{count}, T_{st}, T_K, T_{up}, T_{uplim}$ on $\mathbb{R}$.

Definitions:

Discrete topology on X: is defined by letting every subset of X be open (and hence also closed), and X is a discrete topological space if it is equipped with its discrete topology;

Co-discrete topology:= {∅, X} defines a topology on X,

Finite topology n := {∅} ∪ {A ⊆ X | X \ A is finite} defines a topology on X,

Countable topology:= {∅} ∪ {A ⊆ X | X \ A is countable} defines a topology on X,

Standard topology:=The collection $B_{st}$ := {(a, b) | a < b} defines a basis for a topology on R, called the standard topology,

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

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

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

For each topology,

(i) say if K := {$\frac{1}{n+1}$| n ∈ N} is closed.

(ii) say if K ∪ {0} is closed.

(iii) say if R is Hausdorff.

(iv) say if R is connected, except for the topology $T_{K}$. You can assume that ($\mathbb{R},T_{st})$ is connected, which will be proven in class.

(v) say if Q, endowed with the subspace topology, is connected.

Justify each answer. To make the exercise shorter, it is useful to keep in mind that these topologies are related.


My thoughts so far: In class, we discovered the following relations:

$T_{codisc} \subset T_{count} \subset T_{fin} \subset T_{st} \subset T_K \subset T_{uplim} \subset T_{disc},$ and $ T_{codisc} \subset T_{up} \subset T_{st},$ (not entirely sure whether the place for $\subset T_{count}$ is correct though, please comment ) I think as long as I find one topology where for instance, its left topology is closed and its right topology is not closed, I can assume all topologies on the left is closed and the same for its right side, since the topologies is subset of one another.

(i) take the limit of $\frac{1}{n+1}$, and it approaches to 0. It's open under $T_{dics}$ since each singleton is open. If x < 0, then the open set (3x/2, x/2) is a neighborhood of x that does not intersect K. If x > 1, then the open set $(x-\frac{1-x}{2}, x+ \frac{1+x}{2}$) is a neighborhood of x that does not intersect K,. If x ∈ (0, 1), then there is a natural number n such that $\frac{1}{n+1}< x <\frac{1}{n}$. This interval is a neighborhood of x that does not intersect K. So it's open.

Then I test $T_{k}$, since $subset T_{st} \subset T_K$, and none of the sets we added to the basis intersect with K, so there now are open sets S containing 0 with O∩K=∅∩K=∅.Thus, $T_{k}$ is still closed.

So I think to the left of $T_{st}$ are open, and to its right is closed.

ii) Since now the set includes $0$, I think $T_{st}$ will be closed, with its right side being closed and left side being open.

And I have no clue how to start iii, iv, and v.

Thank you in advance for the help, any comment is greatly appreciated!

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closed as too broad by Xander Henderson, Mohammad Riazi-Kermani, Xam, Mostafa Ayaz, Strants Feb 21 '18 at 16:52

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Define $T_{disc}, T_{codisc}, T_{fin}, T_{count}, T_{st}, T_K, T_{up}, T_{uplim}$ on $\mathbb{R}$. $\endgroup$ – William Elliot Feb 20 '18 at 22:37
  • $\begingroup$ @WilliamElliot the question is updated, is this what you mean? $\endgroup$ – Liz Feb 21 '18 at 1:34
  • $\begingroup$ As in what each topology is so I don't have to look at several Wiki pages. $\endgroup$ – Sean Roberson Feb 21 '18 at 1:39
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    $\begingroup$ I have not seen some of these terms before. The anti-discrete, or coarse topology on a set S is $ \{S.\emptyset\}.$ In the co-countable (or co-finite) topology s non-empty set is open iff its complement is countable (or finite). I have never heard of the upper topology although I may have seen it under another name. What is the K-topology? The upper limit topology on $\Bbb R$ is homeomorphic to the lower limit topology, which is known as the Sorgenfrey line. Please apply the advice in the comment above by Sean Robertson. It will greatly improve your odds of getting answers. $\endgroup$ – DanielWainfleet Feb 21 '18 at 1:52
  • $\begingroup$ Sorry about the confusion, now the definition is added. $\endgroup$ – Liz Feb 21 '18 at 2:09
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Relations between 8 topologies on $\mathbb{R}$

  1. $\mathcal{T}_\mathrm{codisc}\subset\mathcal{T}_\mathrm{fin}\subset\mathcal{T}_\mathrm{st}\subset\mathcal{T}_\mathrm{K}\subset\mathcal{T}_\mathrm{uplim}\subset\mathcal{T}_\mathrm{disc}$
  2. $\mathcal{T}_\mathrm{fin}\subset\mathcal{T}_\mathrm{count}\subset\mathcal{T}_\mathrm{disc}$
  3. $\mathcal{T}_\mathrm{codisc}\subset\mathcal{T}_\mathrm{up}\subset\mathcal{T}_\mathrm{st}$

(i) Say if $K=\{\frac{1}{n}\mid n\in\mathbb{N}\}\subset\mathbb{R}$ is closed.

Claim A: Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$ such that $\mathcal{T}\subset\mathcal{T}'$. Then $A\subset X$ is open (resp. closed) in $\mathcal{T}$ $\Rightarrow$ $A$ is also open (resp. closed) in $\mathcal{T}'$.

As you pointed out, $K$ is closed in $\mathcal{T}_\mathrm{K}$ so that it is also closed in $\mathcal{T}_\mathrm{uplim}\subset\mathcal{T}_\mathrm{disc}$. Moreover, it is closed in $\mathcal{T}_\mathrm{count}$ since $K$ itself is countable.

However, $K$ is neither open nor closed in $\mathcal{T}_\mathrm{st}$ since the limit point $\{0\}$ is not included in $K$. The same is true for $\mathcal{T}_\mathrm{codisc}\subset\mathcal{T}_\mathrm{fin}$ and $\mathcal{T}_\mathrm{up}$.

(ii) Say if $K\cup\{0\}$ is closed.

As you pointed out, $K\cup\{0\}$ is closed in $\mathcal{T}_\mathrm{st}\subset\mathcal{T}_\mathrm{K}\subset\mathcal{T}_\mathrm{uplim}\subset\mathcal{T}_\mathrm{disc}$. Moreover, it is closed in $\mathcal{T}_\mathrm{count}$ since $K\cup\{0\}$ is countable too.

It is easy to see that $K\cup\{0\}$ is not closed in $\mathcal{T}_\mathrm{codisc}\subset\mathcal{T}_\mathrm{fin}$ and $\mathcal{T}_\mathrm{up}$.

(iii) Say if $\mathbb{R}$ is Hausdorff.

Claim B: Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$ such that $\mathcal{T}\subset\mathcal{T}'$. Then $X$ is Hausdorff in $\mathcal{T}$ $\Rightarrow$ $X$ is also Hausdorff in $\mathcal{T}'$.

At least we know that $\mathbb{R}$ is Hausdorff in $\mathcal{T}_\mathrm{st}$ so that it is also Hausdorff in $\mathcal{T}_\mathrm{K}\subset\mathcal{T}_\mathrm{uplim}\subset\mathcal{T}_\mathrm{disc}$.

However, $\mathbb{R}$ is not Hausdorff in any topology $\mathcal{T}_\mathrm{codisc}$, $\mathcal{T}_\mathrm{fin}$, $\mathcal{T}_\mathrm{count}$, $\mathcal{T}_\mathrm{up}$. (It's a good exercise for students who study general topology for the first time.)

(iv) Say if $\mathbb{R}$ is connected.

Claim C: Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on $X$ such that $\mathcal{T}\subset\mathcal{T}'$. Then $X$ is disconnected in $\mathcal{T}$ $\Rightarrow$ $X$ is also disconnected in $\mathcal{T}'$. That is, $X$ is connected in $\mathcal{T}'$ $\Rightarrow$ $X$ is also connected in $\mathcal{T}$.

Assume that $\mathbb{R}$ is connected in $\mathcal{T}_\mathrm{st}$ so that it is also connected in $\mathcal{T}_\mathrm{codisc}$, $\mathcal{T}_\mathrm{fin}$, $\mathcal{T}_\mathrm{up}$.

Moreover, it is trivial to see that $\mathbb{R}$ is disconnected in any topology $\mathcal{T}_\mathrm{uplim}$ and $\mathcal{T}_\mathrm{disc}$.

As you want we skip the case of $\mathcal{T}_\mathrm{K}$. (Looks hard but challenging for students.)

It remains to show that $\mathbb{R}$ is connected in $\mathcal{T}_\mathrm{count}$. Note that every nonempty closed set (except for $\mathbb{R}$) in this topology is countable by definition. On the other hand, every nonempty open set is uncountable. Therefore, $\mathbb{R}$ is connected in this topology since there is no nonempty set which is both open and closed (except for $\mathbb{R}$).

(v) Say if $\mathbb{Q}$, endowed with the subspace topology, is connected.

We know that $\mathbb{Q}$ is disconnected in $\mathcal{T}_\mathrm{st}$ (or we leave the proof for students) so that it is also disconnected in $\mathcal{T}_\mathrm{K}\subset\mathcal{T}_\mathrm{uplim}\subset\mathcal{T}_\mathrm{disc}$.

Moreover, $\mathbb{Q}$ is disconnected in $\mathcal{T}_\mathrm{count}$ since $\mathbb{Q}$ is discrete in this subspace topology.

However, $\mathbb{Q}$ is connected in $\mathcal{T}_\mathrm{fin}$. Note that every nonempty closed set (except for $\mathbb{Q}$) in this topology is finite by definition. On the other hand, every nonempty open set is infinite. Therefore, there is no nonempty set which is both open and closed (except for $\mathbb{Q}$).

It is also true that $\mathbb{Q}$ is connected in $\mathcal{T}_\mathrm{up}$. Note that every nonempty closed set (except for $\mathbb{Q}$) in this topology is bounded above but not bounded below by definition. On the other hand, every nonempty open set (except for $\mathbb{Q}$) is bounded below but not bounded above. Therefore, there is no nonempty set which is both open and closed (except for $\mathbb{Q}$).

It is trivial to see that $\mathbb{Q}$ is connected in $\mathcal{T}_\mathrm{codisc}$.

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  • $\begingroup$ I'm really new to topology, could you explain (iii) a little bit more? I'm not so clear why $T_{st}$ is Hausdorff and also could you proof that $T_{st}$ is connected? Thank you a lot! $\endgroup$ – Liz Feb 21 '18 at 4:20
  • $\begingroup$ @Liz You mentioned that "You can assume that $(\mathbb{R},\mathcal{T}_\mathrm{st})$ is connected, which will be proven in class". It's not simple, and almost all books may contain the proof. On the other hand, it is easy to see that $\mathbb{R}$ is Hausdorff. For any two points $a<b$, we can choose two open sets $a\in(-\infty, \frac{a+b}{2})$ and $b\in(\frac{a+b}{2},\infty)$. $\endgroup$ – ChoF Feb 21 '18 at 4:35

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