this is a topology question:

For these topologies $T_{disc}, T_{codisc}, T_{fin}, T_{count}, T_{st}, T_K, T_{up}, T_{uplim}$ on [0, 1]:

i) find the connected component of $\frac{1}{2}$ in [0, 1], except for $T_K$.

ii) say if [0, 1] is compact. You can assume ([0, 1],Tst) is compact.

iii) say if K := {$\frac{1}{n+1}$| n ∈ N} is compact.


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)} defines a basis for a topology on R, called the standard topology,

K topology := The collection $B_{K} := B_{st}$ ∪ {(a, b) \ K}, 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]} defines a basis for a topology on R.

Here is what I have:

i) I'm not sure how to do this. If I have to give an answer, then for $T_st$, I think we can only have singletons {$\frac{1}{2}$}, as its own connected components? or {$\frac{1}{2}$} $\cup$ {some singleton}. Because $T_st$ defines the interval (a,b), and it's impossible to find an closed interval that fulfills the requirement of connected space. For instance, take the interval (0,$\frac{1}{2}$) and assume it's connected. We can rewrite it as (([0,$\frac{1}{4}$))∪([$\frac{1}{4}$,$\frac{1}{2}$]), since $\frac{1}{4}$∈([$\frac{1}{4}$,$\frac{1}{2}$])≠ϕ , it follows that [0,$\frac{1}{4}$)=ϕ. Therefore, if c∈(0,$\frac{1}{2}$), then c∉[0,$\frac{1}{4}$), and c is greater than or equal to $\frac{1}{4}$. And we can make c≤a because it's arbitrary, thus a=c, and the only connected part is singletons.

Please correct me if I'm wrong, and I don't know how to work this for $T_{disc}, T_{codisc}, T_{fin}, T_{count}, T_K, T_{up}$ (If the above is correct, I think $ T_{uplim}$ will follow something similar.)

ii) Given: Let (X, T ) be a space, and Y ⊆ X a subset. • The space (Y, $T_Y$ ) is compact if and only if any covering of Y by open subsets in X contains a finite subcollection covering Y . • If Y ⊆ Z ⊆ X, (Y, $T_Y$ ) is compact if and only if (Y,($T_Z)_Y$ ) is compact.

For $T_K, T_{up}, T_{uplim}$, I not so sure how to find the finite covering, the idea is too abstract for me. I guess for $T_K$, I think K := {$\frac{1}{n}$} is finite and intuitively, the incountable $\frac{1}{n}$ will cover the entire interval, except at point 0 because it is a limit point.

iii) Similarly, here it's more direct that at 0, we encounter a special point, but I'm still not sure about how the argument should go.

Thank you in advance, any help is appreciated.

  • $\begingroup$ For (i), we can write the interval $[0,1]$ as the disjoint union of its connected components, i.e. as a union of disjoint closed sets, so I would suggest you try to examine what a closed subset of $[0,1]$ is in each topology. For example, $[0,1]$ is connected in the standard topology $T_{st}$, so it has exactly one connected component, i.e. $[0,1]$ itself. On the other hand, in the discrete topology, every point is closed, so $\{\frac{1}{2}\}$ is the connected component containing $\frac{1}{2}$ $\endgroup$ – An Coileanach Feb 28 '18 at 17:13
  • $\begingroup$ @AnCoileanach I see what you mean, this makes a lot sense! Could you take a look at the other questions? $\endgroup$ – Sophie0502 Feb 28 '18 at 18:18
  • $\begingroup$ For (ii), proving compactness relies on open sets being "sufficiently big", so for example $[0,1]$ will fail to be compact in the discrete topology. Have a think about the others, but the following questions may be of some use: math.stackexchange.com/questions/2155720/… and math.stackexchange.com/questions/2017294/… $\endgroup$ – An Coileanach Feb 28 '18 at 20:56
  • $\begingroup$ For (iii), this will answer your question for the standard topology: math.stackexchange.com/questions/1938498/…. For the discrete and co-discrete, you can use similar logic to (i) and (ii). I suspect that there should be some distinction between the cofinite and cocountable cases, since $K$ is countably infinite, but unfortunately, I'm not so sure. $\endgroup$ – An Coileanach Feb 28 '18 at 21:20
  • $\begingroup$ For (i), you may find an answer from Find the largest connected components of $\frac{1}{2}$ in $[0,1]$ for $T_{fin}$, $T_{count}$, $T_{up}$, $T_{uplim}$. $\endgroup$ – ChoF Mar 3 '18 at 6:31

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