# Find the largest connected components of $\frac{1}{2}$ in [0, 1] for $T_{fin}, T_{count}, T_{up}, T_{uplim}$

here is my question, I saw a similar post already, but there's no answer/discussion yet, and I already have something, so pease help me out:

Find the largest connected components of $\frac{1}{2}$ in [0, 1] for $T_{fin}, T_{count}, T_{up}, T_{uplim}$ on [0, 1].

I already proved that $T_{st}$ has the interval [0,1] as connected, and $T_{fin} \subset T_{st}$ and $T_{up}\subset T_{st}$, can I just say due to this inclusion, [0,1] is also the largest connected components in $T_{fin}$ and $T_{up}$? If not, how can I find the connected components for $T_{fin}$ and $T_{up}$?

From the proposition, it says:

"(X, $T_{fin}$) is connected if and only if X is not finite.

(X, $T_{count}$) is connected if and only if X is not countable"

If I want to prove along this proposition, I'm confused when [0,1] is finite and when it's not countable. The intuition is that for both $T_{fin}$ and $T_{count}$, the answer should just be the singleton {1/2}, am I right?

Any help is appreciated!

Definitions: 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,

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

## 1 Answer

I already proved that $T_{st}$ has the interval $[0,1]$ as connected, and $T_{fin} \subset T_{st}$ and $T_{up}\subset T_{st}$, can I just say due to this inclusion, $[0,1]$ is also the largest connected components in $T_{fin}$ and $T_{up}$? If not, how can I find the connected components for $T_{fin}$ and $T_{up}$?

Let $\mathcal{T}\subset\mathcal{T}'$ be two topologies on $X$. Then the identity function $\mathrm{id}_X\colon (X,\mathcal{T}')\to (X,\mathcal{T})$ is always continuous. So if $A$ is connected in $(X,\mathcal{T}')$, then $A=\mathrm{id}_X(A)$ is also connected in $(X,\mathcal{T})$.

Since $[0,1]$ is connected in $T_{st}$ and $T_{fin}\subset T_{st}$, $T_{up}\subset T_{st}$, we conclude that $[0,1]$ is also connected in $T_{fin}$ and in $T_{up}$.

If I want to prove along this proposition, I'm confused when $[0,1]$ is finite and when it's not countable. The intuition is that for both $T_{fin}$ and $T_{count}$, the answer should just be the singleton {1/2}, am I right?

The set $[0,1]$ is neither finite nor countable. It is uncountable. By the propositions above, $[0,1]$ is connected in $T_{fin}$ and in $T_{count}$. So the answer should be $[0,1]$ itself in $T_{fin}$ and $T_{count}$.

For $T_{uplim}$, you can find the answer at Find the connected components of $[0,1]$ with respect to the lower-limit topology $T[,)$.

• Thank you again for the help :) – Liz Mar 1 '18 at 14:39