Usual and order topology coincides

Question

Show that usual and order topology on $\mathbb R$ coincides. And what does the word "coincides" mean in the statement?

Answer 1

If $(X,<)$ is any linearly ordered set, then the order topology $\mathcal{T}_{<}$ is the smallest topology that makes all sets of the form $L_p = \{x \in X: x < p\}$ and $R_p = \{x \in X: x > p\}$, (where $p \in X$) open. (these sets form a so-called subbase for this topology).

If $(X,d)$ is a metric space the metric topology the the smallest topology $\mathcal{T}_d$, such that all open balls $B(x,r) = \{y: d(x,y) < r\}$, (where $x \in X, r \in \mathbb{R}, r>0$) are open. (these then form a base for the metric topology).

NOw the claim is that for the reals $\mathbb{R}$ with their usual order $<$ and the udual metric $d(x,y) = |x - y|$, we have $\mathcal{T}_d = \mathcal{T}_{<}$.

Show this by showing two inclusions.

Hint: use that $B(x,r) = (x-r, x+r) = L_{x+r} \cap U_{x-r}$ e.g. so $\mathcal{T}_d \subseteq \mathcal{T}_{<}$. For the other way round, show that $L_p$ is metric open: for every $y < p$, set $r(y) = p -y$, then $B(y,r(y) \subset L_p$, so every $y \in L_p$ is an interior point in the metric topology, and ditto for $R_p$ as well.