Proof verification that standard topology is a topology

Let $$\beta$$ be the collection of all open intervals $$(a,b) \subset \mathbb{R}$$, with $$a,b \in \mathbb{R}$$. Prove that the topology $$\tau_\beta$$ generated by $$\beta$$ is in fact a topology. Observation:

$$\tau_\beta = \{U \subset \mathbb{R} \ \vert \ \forall x \in U ,\exists B \in \beta \text{ such that } x \in B \subset U\}$$

My proof: $$\emptyset \in \tau_\beta$$ vacuously. Since for every $$x \in \mathbb{R}$$ and $$\varepsilon > 0$$, we have that $$x \in (x-\varepsilon, x+\varepsilon) \subset \mathbb{R}$$, then $$\mathbb{R} \in \tau_\beta$$. Now, given any arbitrary $$x \in A = \displaystyle{\bigcup_{j \in J} U_j}$$, where $$U_j$$ is an open set for each $$j$$, then $$x \in U_i$$ for some $$i \in J$$. Since $$U_i$$ is open, then there exists an interval $$B$$ such that $$x \in B \subset U_i$$. Finally, consider $$W = U_1 \bigcap U_2$$, where $$U_1$$ and $$U_2$$ are open sets. Given any $$x \in W$$, since $$x \in U_1$$ and $$x \in U_2$$, then there exist $$B_1, B_2 \in \beta$$ such that $$x \in B_1 \subset U_1$$ and $$x \in B_2 \subset U_2$$. Then $$x \in B_1 \bigcap B_2 \subset U_1 \bigcap U_2$$ and we're done (the rest is trivial by induction).

Are all my steps correct?

• Yes it's correct. Maybe you should spell out the observation in the beginning.. – Berci Jan 7 at 22:48
• Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from? – Chris Custer Jan 8 at 3:42
• This was an exercise given on my first day of a general topology summer course. – Matheus Andrade Jan 8 at 10:15
• Ok. I guess it gets you to work through the definitions. – Chris Custer Jan 8 at 19:31

show $$\cap$$T is a topology for X.