Showing collection of subsets of R is a topology

general-topology

Show that the collection of subsets of $\Bbb R$

$\mathscr T =\{(a,+\infty)|a ∈ \Bbb R\} ∪ \{\Bbb R\} ∪ \{\emptyset\}$
is a topology i.e. a runs over all real numbers, i.e. for every a we have $(a,+\infty) \in \mathscr T $.

Show that $( \Bbb R, \mathscr T )$ is not a Hausdorff topological space

So I think I should be basing my answer for first part like part c from Prove that the following collection of subsets of $\Bbb R$ is a topology? but i'm not really sure, can anyone give any hints to help?

Best Answer

Let $\mathscr T =\{(a,+∞)|a ∈ \Bbb R \} ∪ \{\Bbb R\}∪\{\emptyset\}$

  1. $\emptyset$, $\Bbb R \in \mathscr T$ by definition
  2. Let $\{S_i\}_{i \in I}$ with $S_i \in \mathscr T$. If $\exists i \ $ s.t. $S_i=\Bbb R$ then $\cup_{i \in I}S_i=\Bbb R \in \mathscr T$. If $S_i\neq \Bbb R \ \forall i$ then $S_i=(a_i, +\infty)$ and so $\cup_{i \in I}S_i= (\inf a_i, +\infty) \in \mathscr T$ ($\inf$ can be $-\infty$
  3. Let $S_1, S_2 \in \mathscr T$. If $S_i=\mathbb R$ or $S_i=\emptyset$ $S_1 \cap S_2 \in \mathscr T$ trivially. Else $S_i=(a_i, +\infty)$ and $S_1 \cap S_2=(\max a_i,+\infty) \in \mathscr T$

$\mathscr T$ is not $T_2$: let $x,y\in \Bbb R$ s.t. $x < y$. Let $A \in \mathscr T$ an open not empty set s.t. $x\in A$. If $A= \Bbb R$ then $y \in A$ if $A=(a, +\infty)$ then $y\in A$. So you can't find $A,B$ disjoint open neighborhoods of $x$ and $y$

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