Consider $T_s=\{∅\}\cup\{U\in P(\mathbb R)\mid ∀ x\in U,∃\epsilon\gt0:[x,x+\epsilon)\subset U\}$. Prove that $T_s$ is a topology over $\mathbb R$ In $\mathbb R,$ consider the topology $T_s=\{\emptyset\}\cup\{U\in P(\mathbb R)\mid \forall x\in U,\exists\epsilon\gt0:[x,x+\epsilon)\subset U\}$. Prove that $T_s$ is a topology over $\mathbb R$.
Let's see if $T_s$ is closed under $\cup.$
Let $\{U_\alpha:\alpha\in I\}\subset T_s.$ 
Let $x\in\bigcup_{\alpha\in I}U_\alpha $. We want to show that $x\in T_s$
As $x\in\bigcup_{\alpha\in I}U_\alpha, $ then $\exists\alpha_0\in I: x\in U_{\alpha_0}$.
But $x\in U_{\alpha_0}\subset \{U_\alpha:\alpha\in I\}\subset T_s$. Thus $x\in T_s$
Now let's see if $T_s$ is closed under $\cap.$
Let $\{U_i:i\in I\}\subset T_s,$ where $I$ is finite.
Suppose $U_i\neq \emptyset,\forall i\in I$.
Let $x\in \bigcap U_i$, then $x\in U_i,\forall i$.
But $x\in U_i\subset \{U_i:i\in I\}\subset T_s$. Thus $x\in T_s$.
Please let me know if you find my proof correct. 
 A: 
Let's see if $T_s$ is closed under $\cup.$
Let $\{U_\alpha:\alpha\in I\}\subset T_s.$ 
Let $x\in\bigcup_{\alpha\in I}U_\alpha $. We want to show that $x\in T_s$

That's not right; what you'd want to show is that $\bigcup_{\alpha\in I} U_\alpha \in T_s,$ not that a member of $\bigcup_{\alpha\in I} U_\alpha$ is in $T_s.$
And the same applies to intersections.
A: You are trying to prove the wrong thing. The topology axioms are:

Let $X$ be a set, a topology on $X$ is a collection $\tau$ of subsets of $X$ such that:

*

*$X,\emptyset\in \tau$,

*For every arbitrary collection $\{U_{\lambda }\in \tau : \lambda \in \Lambda\}$ we have $\bigcup_{\lambda\in \Lambda}U_\lambda \in \tau$

*For every finite collection $U_1,\dots,U_k\in \tau$ we have $U_1\cap \cdots \cap U_k\in \tau$.


For this case (1) is trivialy verified. What you seem to be missing is that elements of the set $\tau$ are subsets of $X$. So a topology is a set of sets.
When you say that you must show that given $\{U_\alpha : \alpha \in I\}\subset T_s$ we must show that $x\in \bigcup_{\alpha\in I}U_\alpha$ implies $x\in T_s$ you are confusing things so it is important to point this out. For starters, you can't even have $x\in T_s$ because elements of $T_s$ are subset of $\mathbb{R}$ and not numbers.
Now, how to do it correctly? Given $\{U_\alpha : \alpha \in I\}\subset \tau$ you need to show that $U=\bigcup_{\alpha\in I}U_\alpha\in \tau$. What it means to show that $U\in \tau$? It is either empty or satisfy the condition you mention. If it is empty there is nothing to prove. If it is not empty then you must show for all $x\in U$ there is $\epsilon > 0$ with $[x,x+\epsilon)\subset U$.
That is what must be shown in the case of the union and similarly for the intersection. Now, if $x\in U$ there is $\lambda$ with $x\in U_\lambda$. If $x\in U_\lambda$ since $U_\lambda\in \tau$ it must be the case that there is $\epsilon > 0$ such that $[x,x+\epsilon)\subset U_\lambda$. But $U_\lambda\subset U$, hence $[x,x+\epsilon)\subset U$. Consequently for every $x\in U$ there is $\epsilon > 0$ with $[x,x+\epsilon)\subset U$ and (2) is verified.
For three you do the same. Set $U=U_1\cap \cdots \cap U_k$. If $x\in U$ then $x\in U_i$ for all $i=1,\dots,k$. Since $U_i\in\tau$ there is $\epsilon_i>0$ with $[x,x+\epsilon_i)\subset U_i$.
Take $\epsilon = \min\{\epsilon_i\}$. Then since $\epsilon < \epsilon_i$ it follows that $x + \epsilon < x+\epsilon_i$ and thus since we are dealing with intervals $[x,x+\epsilon)\subset [x,x+\epsilon_i)$.
But $[x,x+\epsilon_i)\subset U_i$ for all $i$, hence $[x,x+\epsilon)\subset U_i$ for all $i$ and by the very definition of the intersection we have $[x,x+\epsilon)\subset U$. Hence in accord to the definition of open sets you gave, (3) is verified.
