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In a metric space $E$, how to prove that the connected component of a point $x\in E$ is contained in every open and closed set containing $x$.

Roiner Segura Cubero
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  • What is a good description of connected component? Is it closed? Is it open? – Sigur Jun 03 '13 at 16:56
  • $C(x)$ the connected component of $x$ in E, is the union of all connected subsets of $E$ containing a point $x\in E$. $C(x)$ is the largest connected set containing $x$. Thanks Sigur!! – Roiner Segura Cubero Jun 03 '13 at 17:07
  • Just to clarify: you want to prove that it is contained in every set which is both open and closed, right? Some authors write clopen. – Sigur Jun 03 '13 at 17:12
  • Yes Sigur, I want to prove that it is contained in every set which is both open and closed. – Roiner Segura Cubero Jun 03 '13 at 17:19
  • What are the clopen subsets of a connected space? – Sigur Jun 03 '13 at 17:20
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    the unique "clopen" sets of a connected set are the empty set and itself, but as this "clopen" set contained $x$ then is $C(x)$ y therefore $C(x)$ is contained in every open and closed set containing x. Right? – Roiner Segura Cubero Jun 03 '13 at 17:31

1 Answers1

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The fact that $E$ is metric is really irrelevant, what matters is that it is a topological space.

Let $x \in E$ and let $C(x)$ be the connected component of $x$. If $O$ is closed and open ("clopen") and $x \in O$, then, as $O$ and $E \setminus O$ are both open, $C(x)$ cannot intersect $O$ and $E \setminus O$ in a non-empty set, or else

$$ C(x) = ( C(x) \cap O ) \cup ((C(x) \cap (E \setminus O)) $$

would be a non-trivial decomposition of the connected set $C(x)$ by relative open sets. So one intersection is empty, and it cannot be $C(x) \cap O$, as it contains $x$. Alternatively stated: $C(x) \cap O$ is clopen in $C(x)$, and non-empty so must be $C(x)$ by connectedness.

So $C(x) \cap O = C(x)$, or $C(x) \subset O$.

This shows that $C(x)$ is contained in any clopen that contains $x$.

(the intersection of all clopen sets that contain $x$ (there is always at least one: $E$) is called the pseudo-component of $x$, so this proves that the component of $x$ is always a subset of the pseudocomponent of $x$; they can differ in general.)

Henno Brandsma
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