One of the definition of basis of a topology is that any open set can be expressed as unions of basis, but I always thought the other definition was easier to use. So for this definition, how do we express $(e, \pi)$ as unions of open intervals with rational endpoints which forms a basis for the usual topology for R.
How do we express $(e, \pi)$ as unions of open intervals with rational endpoints
general-topologymetric-spaces
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Let $B = \mathcal{I} \cup \mathcal{J}$ where $\mathcal{I} = \left\{ \mathbb{R} \right\}$, $\mathcal{J} = \left\{ (a,b) \cap \mathbb{Q} | a<b, a,b \in \mathbb{Q} \right\}$. Suppose that $U,V \in B$. If $U,V \in \mathcal{I}$, then clearly $U \cap V \in \mathcal{I}$. If one of $U,V$ is in $\mathcal{I}$ whereas the other is in $\mathcal{J}$, then clearly $U \cap V \in \mathcal{J}$. If both $U$ and $V$ are in $\mathcal{J}$, then $U \cap V \in \mathcal{J}$ from the basic properties of open intervals. Hence $B$ forms a basis.
Let $\mathcal{T}_B$ be the topology generated by $B$ and $\mathcal{T}$ be the usual topology. Since the interval $(0,1) \in \mathcal{T}$ contains irrational numbers $(0,1) \not \in \mathcal{T}_B$. Therefore $\mathcal{T}_B$ does not contains $\mathcal{T}$. Since the interval $(0,1) \cap \mathbb{Q} \in \mathcal{T}_B$ is not open in the usual topology, we have $\mathcal{T}$ does not contains $\mathcal{T}_B$.
The space $X = (\mathbb{R}, \mathcal{T}_B)$ is compact. To see this let $\mathcal{O}$ be an open cover of $X$. Since $\sqrt{2} \in U \in \mathcal{O}$ implies $U = \mathbb{R}$, we have a finite subcover $I \subseteq \mathcal{O}$.
The space $X$ is separable. To see this, consider $D = \left\{ 0 \right\}$. For every irrational $r$, there is unique neighborhood $\mathbb{R}$ showing that the closure of $D$ is just $X$. Hence $X$ is separable.
The space $X$ is connected. To see this, suppose that $U$ and $V$ are proper non empty open subsets of $X$. Then they cannot cover $X$ simply because they do not have any irrational numbers.
The ordered square is the set $[0,1] \times [0,1]$ in the order topology from the order that the square inherits from $\mathbb{R} \times \mathbb{R}$ with the lexicographic order (the lexicographically ordered plane).
Munkres already spent some time warning the reader that this is not the same topology that $[0,1] \times [0,1]$ gets as a subspace of the ordered plane.
In general, if $(X,<)$ is an ordered set with order topology $\mathcal{T}_<$ and $A \subseteq X$ we can consider two topologies on $A$: the subspace topology $\mathcal{T}_<|_A = \{O \cap A: O \in \mathcal{T}_<\}$ that $A$ inherits from $(X,\mathcal{T}_X)$ and the order topology that $A$ gets when we restrict the order first to $A$ and then generate the topology from that restricted order.
And these topologies concide when $A$ is order-convex: $\forall a_1 \in A, a_2 \in A: \forall b \in X: (a_1 < b < a_2 ) \to b \in A$.
Even in the reals you can see this phenomenon: Take $A = [0,1) \cup \{2\}$. As a subspace of the reals (ints order topology), $2$ is an isolated point, i.e. $\{2\}$ is open, while in the restricted order topology $2$ lies "just above" $[0,1)$ and in the restricted order topology the set is homeomorphic to $[0,1]$. Note that $A$ is not order convex.
Also $A = [0,1]\times [0,1]$ is not order convex in the ordered plane: e.g. $(0,1), (1, 0)$ are both in $A$, while $(0,2)$ which lies between them lexicographically is not.
Munkres does take the square in its (restricted) order topology, not the subspace topology, so the subbase for its topology (as in any order topology) is the set of all open segments both up and down, and the derived base has the two special cases involving $(0,0) = \min(A)$ and $(1,1) = \max(A)$. So not only open intervals, also sets like $[(0,0), (0,t))$ and $((1,t), (1,1)]$ are basic open.
Best Answer
There is no real problem: consider the sets $A=\{a\in\mathbb{Q}:e<a<(e+\pi)/2\}$ and $B=\{b\in\mathbb{Q}:(e+\pi)/2<b<\pi\}$ Then $$ (e,\pi)=\bigcup_{\substack{a\in A \\ b\in B}}(a,b) $$ Clearly the union is contained in $(e,\pi)$. The infimum $m$ of the union has to be $e$, otherwise there would be $r\in\mathbb{Q}$ such that $e<r<m$ (between any two real numbers there is at least one rational number). This contradicts $m$ to be the infimum, because $r\in A$.
Similarly, the supremum of the union is $\pi$ and so the union is the whole $(e,\pi)$.
We used no particular property of $e$ and $\pi$: indeed the same argument proves that any open interval is the union of open intervals with rational extremes.