Give an example of a topological space $X$, and a connected component which is not open in $X$

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Give an example of a topological space $X$, and a connected component which is not open in $X$

I know of the following theorems:

  • Each connected component is closed;
  • If a topological space has a finite number of connected components, then those components are open in $X$.

Because of the second theorem, I need to be on the lookout for a topological space with a infinite number of connected components. I was thinking of something like this:

$$X = \bigsqcup_{n\in \mathbb{N}} \left[n, \dfrac{3}{2} n\right[$$

But this doesn't seem to work since $\sqcup_{n\geqslant 2} [n, \frac{3}{2}[$ as an union of infinite closed sets seems to be closed again.

The open/closed concept in a relative topology can be mindbending. Could someone help me to find an simple example?

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The space $\displaystyle X = \Big \{\frac{1}{2}, \frac{1}{3}, \ldots, 0 \Big\}$ is completely disconnected. That is to say the connected components are the singletons $\displaystyle \Big \{\frac{1}{2}\Big\}, \Big \{\frac{1}{3}\Big\} \ldots $ all of which are open, and $\{0\}$ which is not.

To show $\{0\}$ is not open observe the series $\frac{1}{2}, \frac{1}{3}, \ldots$ converges to zero, so is eventually in any open set about $0$. Therefore $\{0\}$ fails to be open. To show $\{0\}$ is a component observe that for each $n$ we can write $X=\Big \{\frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n} \Big \} \cup \Big\{\frac{1}{n+1}, \frac{1}{n+2}, \ldots , 0 \Big\}$. This demonstrates that $\displaystyle\frac{1}{n}$ and $0$ are in different components.

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Take $$X=\mathbb{Q}$$ the rationals numbers with the Euclidean topology.