Prove that $\mathcal{T}_1 \subset \mathcal{T}$ with $\mathcal{T}_1$ is the standard topology on $\mathbb{R}$. Let $\mathcal{B}= \left\lbrace [a,b): a,b \in \mathbb{R}, a<b \right\rbrace$. Let $\mathcal{T}_1$ be the standard topology on $\mathbb{R}$
I have already proved $\mathcal{B}$ is a base of a topology $\mathcal{T}$ on $\mathbb{R}$ but how to prove that $\mathcal{T}_1 \subset \mathcal{T}$ ?
 A: We know that $\{(a,b) : a,b \in \mathbb{R}, a<b\}$ is a basis for the standard topology $\mathcal{T}_1$ on $\mathbb{R}$.
For every such set we have $$(a,b) = \bigcup_{\gamma \in (a,b)}[\gamma, b) \in \mathcal{T}$$
Furthermore, every open set $U \in \mathcal{T}_1$ can be written as a union of intervals $(a_i, b_i)$ for $i \in I$ so $$U = \bigcup_{i \in I}(a_i, b_i) \in \mathcal{T}$$
since $\mathcal{T}$ is closed under unions. Therefore $\mathcal{T}_1 \subseteq \mathcal{T}$.
A: I think it's as simple as noting that
$$(a,b) = \bigcup_{p\in(a,b)}[p,b),$$
with a brief argument about why it's enough to show this is true for basis elements, right?
A: A slightly different approach:
Let $O$ be open in $\mathcal{T}_1$. Then for each $x \in O$ we have that $(x-r_x,x+r_x) \subseteq O$ where $r_x >0$.
But then $x \in O$ is an interior point of $O$ in $\mathcal{T}$ too, as $[x,x+r_x) \subseteq O$ too, a fortiori. As $x \in O$ is arbitrary, $O$ is open in $\mathcal{T}$. So $\mathcal{T}_1 \subseteq \mathcal{T}$.
