Let $X$ locally compact, Hausdorff and non compact. Ends. 
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 $K\subseteq K′$.
 A: 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).)
A: 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.
