The space [0,1] relativesed with usual topology is Hausdorff

general-topology

Source: On Minimal Separation Axioms in Topological Spaces (2012)

  • Definition(Minimal Closed Set): A proper nonempty closed subset $F$ of a topological space $X$ is said to be minimal closed if any
    closed set which is contained in $F$ is $\emptyset$ or $F$.

    Definition(Minimal Completely Regular Space): For every point $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in
    F$    , there exist a continuous map $f:X\rightarrow [0,1]$ such that
    $f(x)=0$ and $f(F)=\{1\}$.

    Definition(Minimal c-regular Space): For every point $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in F$, there
    exists open sets $U,V\subset X$ such that $x\in U$, $F\subset V$ and
    $U\cap V=\emptyset$.

Theorem: Every min-completely regular space is min c-regular space.

Proof: Let $X$ be a min-completely regular space. To prove $X$ is min c-regular space. Let $x\in X$ and $F$ be any minimal closed set in $X$ such that $x\not\in F$. Since $X$ is min-completely regular space, there exists a continuous map $f:X\rightarrow [0,1]$ such that $f(x)=0$ and $f(F)=\{1\}$. Also, it is easy to see that the space $[0,1]$ with relativesed usual topology is Hausdorff space. Hence, there exist open subsets $G,H\subset [0,1]$ such that $0\in G$, $1\in H$ and $G\cap H=\emptyset$.

Since $f$ is continuous, $f^{-1}(G),f^{-1}(H)\subset X$ are open such that

$f^{-1}(G)\cap f^{-1}(H)=f^{-1}(G\cap H)=f^{-1}(\emptyset)=\emptyset$,

$f(x)=0\in G \implies f(x)\in G \implies x\in f^{-1}(G)$,

$f(F)=\{1\}\subset H \implies f(F)\subset H \implies F\subset f^{-1}(H)$.

Thus, for every $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in F$, there exists open sets $f^{-1}(G),f^{-1}(H)\subset X$ such that $x\in f^{-1}(G)$, $F\subset f^{-1}(H)$ and $f^{-1}(G)\cap f^{-1}(H)=\emptyset$. Hence, $X$ is min c-regular space.

Now, I want to clarify the sentence "the space $[0,1]$ with relativised usual topology is Hausdorff."

Firstly, I need to know that what the usual topology on $[0,1]$ is.
I have tried to find the base of the usual topology on $[0,1]$ but…

$X=[0,1]$, $\tau=$ Standard Topology

If I consider that the base is $\beta=\{(a,b)\in X : 0<b<a<1\}$, then I cannot generate $[0,1]$.

Thank you for your helps..

Best Answer

This means that $[0,1]$ in the inherited topology from $\Bbb R$ (in the usual topology too) is Hausdorff. This is quite clear as that topology is metric (so Hausdorff), ordered (another reason it's Hausdorff) and hence so are all its subspaces as well as Hausdorff-ness is hereditary. No more than that.

Related Question