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It is known that in cases of locally connected topological spaces or compact Hausdorff spaces the components and quasicomponents coincide. Both claims can be proved using the fact that if a quasicomponent is open, then it is connected.

Does the opposite of the last claim hold too, i.e., if the component is open, does it imply that it must coincide with a quasicomponent?

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Yes. If a component $C$ is open, then it is clopen. It cannot contain a non-empty strictly smaller clopen subset, thus it is the intersection of all clopen subsets containing a fixed point $x \in C$.

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  • $\begingroup$ I see. Thank you! Just for clarification, may I add that since $C$ is a clopen component, if $C$ contained a clopen nontrivial subset, then $C$ would contain a quasicomponent which is not equal to $C$ which is a contradiction. Thus, $C$ cannot contain a non-empty strictly smaller clopen subset, as you have stated. $\endgroup$
    – Emo
    Commented Apr 2, 2023 at 18:18
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    $\begingroup$ @Emo This is correct. An easier argument is this: If $C$ contained a clopen nontrivial subset, then we would have a splitting of the connected $C$ into non-empty disjoint open subsets. $\endgroup$
    – Paul Frost
    Commented Apr 2, 2023 at 21:47

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