Let $X$ be locally compact, Hausdorff and non compact. Prove that if $X$ has one “end”, then $X^\wedge - X$ , (where $X^\wedge$ is any Hausdorff compactification), is a continuum (=compact, connected).

Definition Let $X$ be a topological space. An "end" of $X$ assigns, to each compact subspace $K$ of $X$, a connected component $eK$ of its complement $X\setminus K$, in such a way that $eK′\subseteq eK$ whenever $K\subseteq K′$.

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    $\begingroup$ Well, we know your space X^ is a compact Hausdorff space, and we know that X^-X is compact Hausdorff as well. Moreover, each closed subset of X^-X is closed in X^ as well. Now, if we assume a non-trivial partition of (X^-X) into disjoint open subsets U and V We see first: U and V are closed in (X^-X) as well, hence compact, hence closed in X^. But then X^ is a normal Hausdorff space, and U and V are disjoint closed subsets in X^. I don't know how to achieve a contradiction from here. I'm stucked.. $\endgroup$ – Sara Nein Läufer Dec 10 '15 at 0:03
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    $\begingroup$ @Rise A locally compact space is open in any its Hausdorff compactification (see, for instance, Engelking’s “General topology”, Th. 3.3.9). $\endgroup$ – Alex Ravsky Dec 10 '15 at 8:29
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    $\begingroup$ @Paul: That won’t quite work. Suppose that $X=[0,\to)$. This has only one end: for any compact $K\subseteq X$ we let $eK$ be the unbounded component of $X\setminus K$. But clearly $X\setminus K$ need not be connected; indeed, it can have infinitely many components. $\endgroup$ – Brian M. Scott Dec 10 '15 at 18:46
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    $\begingroup$ @SaraNeinLäufer - Any compact subset $K$ of Brian's $X$ is contained in $[0, a]$ for some $a$. The complement of $[0,a]$ is $(a,\rightarrow)$ so $(a,\rightarrow) = e[0, a] \subseteq eK$. Thus $eK$ must be the unbounded component of $X \setminus K$. $\endgroup$ – Paul Sinclair Dec 10 '15 at 20:42
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    $\begingroup$ If you are willing to risk another untested idea from me: each component $C$ of $\hat X \setminus X$ determines a component of $X \setminus K$. By choosing $eK$ to be the component determined by $C$, you can define an end. If you choose a different component $B$ of $\hat X \setminus X$, that should define a different end. Thus if you have only one end, you can have only one component of $\hat X \setminus X$. $\endgroup$ – Paul Sinclair Dec 10 '15 at 20:48

This isn't true in general. For instance, suppose $X=[0,\infty)\sqcup D$, where $D$ is an infinite discrete space. Then $X$ has only one end (the one coming from $[0,\infty)$, since every connected component in $D$ is compact). But you can compactify $X$ as $X^\wedge=[0,\infty]\sqcup E$ for any compactification $E$ of $D$, and then $X^\wedge-X=\{\infty\}\sqcup (E-D)$ is disconnected.

However, if you assume $X$ is connected and locally connected, then it is true. First, I claim that for any compact set $K\subset X$, $X-K$ has only finitely many unbounded components (where "unbounded" means its closure in $X$ is not compact). To prove this, let $L\subset X$ be a compact set containing $K$ in its interior (such an $L$ exists by local compactness of $X$). Note that by local connectedness, every component of $X-K$ is open, and so by compactness only finitely many components of $X-K$ can intersect $\partial L$. But if $A\subseteq X-K$ is an unbounded component, it is not contained in $L$ and so $A-L$ is a nonempty open set. Since $X$ is connected, $A-L$ cannot be closed in $X$. But $A$ is closed in $X-K\supseteq X-int(L)$, and so $\overline{A-L}$ is both contained in $A$ and must contain points of $\partial L$. Thus $A$ intersects $\partial L$, and by the remarks above, this means there are only finitely many such $A$.

Second, I claim that if $K\subseteq K'\subset X$ are compact subsets and $A$ is an unbounded clopen subset of $X-K$ (in particular, if $A$ is an unbounded component of $X-K$), then $A$ contains an unbounded component of $X-K'$. To prove this, let $L\subset X$ be a compact set containing $K'$ in its interior. As above, only finitely many components of $X-K'$ intersect $\partial L$. Moreover, the argument of the previous paragraph shows that every component of $X-K'$ that intersects $X-L$ must also intersect $\partial L$. We conclude that every component of $A-K'$ is either contained in $L$ or is one of the finitely many components of $X-K'$ intersecting $\partial L$. If all of these finitely many components are bounded, then $A-K'$ would be bounded (since it is contained in the union of $L$ and finitely many bounded sets). This is impossible, since $A$ is unbounded. Thus one of the components of $A-K'$ is unbounded.

It now follows by a standard compactness argument that if $A$ is an unbounded component of $X-K$ for some compact $K\subset X$, then there is an end $e$ of $X$ such that $eK=A$. (Explicitly, let $F_K$ denote the set of unbounded components of $X-K$ with the discrete topology. Then an end is a point in the product $\prod_K F_K$ satisfying certain identities, and paragraph above shows that any finite number of those identities can be satisfied by an element of the product sending $K$ to $A$. Compactness of the product then gives an element satisfying all of the identities.)

Now suppose $X^\wedge$ is a compactification of $X$ such that $X^\wedge-X$ is disconnected. We will show $X$ has more than one end. Let $C$ be a nonempty proper clopen subset of $X^\wedge-X$. Then $C$ and $D=(X^\wedge-X)-C$ are disjoint closed subsets of $X^\wedge$, so we can find disjoint open sets $U,V\subset X^\wedge$ such that $C\subset U$ and $D\subset V$. We then have that $K=X^\wedge-(U\cup V)$ is compact and contained in $X$.

Now $X\cap U$ is an unbounded clopen subset of $X-K$, so as shown above (taking $K'=K$), it must contain an unbounded component of $X-K$. This unbounded component then extends to an end $e$ such that $eK\subset U$. But by the same argument with $V$ in place of $U$, there also exists an end $e'$ such that $e'K\subset V$. We thus have two distinct ends of $X$.

(The argument above was adapted from the standard proof that if $X$ is locally compact Hausdorff, connected, and locally connected, then you can compactify $X$ by adding a point for each end of $X$. This proof can be found here, among many other places (the hypothesis of $\sigma$-compactness used there is easily seen to be unnecessary).)

  • $\begingroup$ Well done. That partly explains why I was spinning my wheels. One question that still remains is whether local-connectedness is necessary. $\endgroup$ – Paul Sinclair Dec 16 '15 at 1:07
  • $\begingroup$ Yeah, I tried for a while to get a connected but not locally counterexample and couldn't come up with one. $\endgroup$ – Eric Wofsey Dec 16 '15 at 1:12

Thank's to everyone, I think I did it. Let me know if you find some mistake.

Let $X$ locally compact, Hausdorff and non-compact. Prove that if $X$ has only one end, then $X^{*} - X$ for any compactification $X^{*}$ (Hausdorff) of $X$ is a continuum (=compact, connected).

Solution Since $X^{*}$ is a Hausdorff compactification of $X$, $X^{*}$ is compact Hausdorff. Now, since $X$ is locally compact, $X$ is open in $X^{*}$, so $X^{c}$ is closed in $X^{*}$, therefore, $K=X^{*}-X= X^{*} \cap X^{c}$ is closed in $X^{*}$ and thereby compact.

Assume that $K$ is disconnected, then there exists an $(U,V)$ separation of $K$ such that $U$ y $V$ are closed, $K=U \cup V$ and $U \cap V=\emptyset$. But $X^{*}$ is normal, since it is compact Hausdorff, so there exists $C$ y $D$ opens of $X^{*}$ such that $U \subseteq C$, $V \subseteq D$ and $C \cap D =\emptyset$. Let $W=X^{*}-(C\cup D)$. We observe that $W \subseteq X$ since, \ $W=X^{*}-(C\cup D) \subseteq X^{*} \cap (U \cup V)^{c}= X^{*} \cap K^{c}=X^{*} \cap (X^{*}-X)^{c}=X^{*} \cap X =X$.\ On the other hand $W=X^{*} -(C \cup D)= X^{*} \cap (C \cup D)^{c}$ is closed, since $(C \cup D)^{c}$ is closed. $X$ is Hausdorff and $W$ is closed, then $W$ is compact in $X$. \ Define $U_{X} = C \cap X$ and $V_{X} = D \cap X$, and observe that they are open in $X^{*}$, (since $X$ is open in $X^{*}$), so they are open in $X$ and therfore $U_{X} \cap V_{X}=\emptyset$. Now let us calculate the complement of $W$ in $X$. We have that \begin{align*} X-W & = X-(X^{*}-(C \cup D))\\ & = X \cap (X^{*} \cap (C \cup D)^{c})^{c}\\ & = X \cap ((X^{*})^{c} \cup (C \cup D))\\ & = (X \cap (X^{*})^{c}) \cup (X \cap (C \cup D))\\ & = (X \cap C) \cup (X \cap D)\\ & = U_{X} \cup V_{X}\\ \end{align*}

So, the complement of $W$ are two disjoint connected components, which contradicts the fact that $X$ has only one end..

Therefore $X^{*} - X$ is a continuum.

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    $\begingroup$ You are okay up to the contradiction (though there are simpler demonstrations for a lot of it), but one compact set having a disconnected complement does not contradict that $X$ has only one end. As Brian mentioned in his example, there are many compact sets in it that have disconnected complements, some with infinitely many components. But his example has only one end. $\endgroup$ – Paul Sinclair Dec 11 '15 at 3:37
  • $\begingroup$ @PaulSinclair So, what can I do? I'm lost then... $\endgroup$ – Sara Nein Läufer Dec 11 '15 at 23:42
  • $\begingroup$ My own attempts have not been successful. I think this is an interesting question, so I've offered a bounty on it. Sorry I've haven't been more helpful. $\endgroup$ – Paul Sinclair Dec 13 '15 at 21:53

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