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Let X be a Go space. If G is open in X, why is every convex component of G open?

( It is well known that any non-void subset G of X can be uniquely represented as a union

of its maximal convex subsets, which are called convex components of G. )

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up vote 2 down vote accepted

Lemma: if $K$ and $L$ are (order) convex in a linearly ordered set $X$, and $x$ is in $K \cap L$, then $K \cup L$ is convex as well.

Proof: suppose $a < b$ are in $K \cup L$ and $c$ lies in $(a,b)$. Consider 2 cases:

1) $c \le x$: then $a$ is in one of the convex sets, say $K$, and so is $x$, but then $x \in (a, x]$ is in $K$ as well, and we are done.

2) $x < c$: then we have the same argument with $x$ and $b$.

Now, if $X$ is a GO-space, $G$ an open subset of $X$, and $K$ a maximal convex subset of $G$, then let $x$ be an arbitrary point of $K$. As $X$ is a GO-space, $x$ has a local base of convex neighbourhoods, so pick a convex neighbourhood $U$ of $x$ such that $U \subset G$. By the lemma, as $K$ and $U$ intersect (in $x$) and are both convex, their union is a convex subset of $G$ and by maximality of $K$, $K \cup U \subset K$, or $U \subset K$, showing $x$ is an interior point of $K$; hence $K$ is open.

Note that the proof is completely analogous to the proof of the fact that in a locally connected space the components of the open sets are open; there we use that the union of intersecting connected sets is connected and the fact that all points have local bases of connected neighbourhoods, here we use the same with connected replaced by convex.

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