say X has finitely many connected components, can we get X locally connected?
I'm thinking to prove this and then I can have every component is open.
Thanks!
X has finitely many connected components, can we get X locally connected
connectednessgeneral-topologylocally-connected
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Best Answer
No: for instance, the space $X=\left\{(x,y)\in\Bbb R^2\,:\, x=0\lor y=\sin\frac1x\right\}$ with the topology inherited from $\Bbb R^2$ is connected, but not locally connected (all sufficiently small balls around $(0,t)$ for $t\in[0,1]$ contain infinitely many cuts of the topologist's sine curve but not a whole arc connecting them).
However, it is true that a topological space with finitely many connected components has all connected components open. This is because the connected components of a topological space are always closed, and therefore each connected component is $C_i=X\setminus\bigcup_{1\le j\le m\\ j\ne i} C_j$, which is in fact the complement of a closed set.