Let $\tau$ be the usual topology on $\mathbb R$. Let $\tau'$ be the topology generated by $\tau \cup \{\mathbb Q\}$. Then is the space $(\mathbb R,\tau')$ connected?

If so, the it will answer the other question I had: Let $(L,<)$ be a linearly ordered set.Let $\tau$ be the order topology on $L$ induced by $<$. Suppose the topological space $(L,\tau)$ compact and connected (i.e., $(L,<)$ is a dense complete linear order with first and last elements). Suppose $\tau'$ is a topology on $L$ which is strictly finer that $\tau$. Is $(L,\tau')$ necessarily disconnected?

  • $\begingroup$ The topology is generated by $\tau\cup\{\Bbb Q\}$, not $\tau\cup\Bbb Q$. $\endgroup$ – Greg Martin Sep 5 '15 at 0:22

Yes, $(\mathbb R,\tau')$ is connected. Notice that all sets $T$ in $\tau'$ can be written in the following form: $$T=A\cup (B\cap \mathbb Q)$$ for $A,B \in\tau$.

Suppose we take any two non-empty and disjoint sets $T$ and $T'$ in $\tau'$. Then $$T=A\cup (B\cap \mathbb Q)$$ $$T'=A'\cup (B'\cap \mathbb Q)$$ Since $\mathbb Q$ is dense, we can conclude that $A\cup B$ and $A'\cup B'$ are disjoint. Since $\mathbb R$ is connected and those two sets are disjoint and open, $(A\cup B) \cup (A' \cup B')$ cannot cover $\mathbb R$. This set contains $T\cup T'$ which can therefore not cover the space.


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