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Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed connected subsets of $X$ that is simply ordered by proper inclusion. Show that $$Y = \bigcap_{A \in \mathcal{A}} A$$ is connected. [Hint: If $C \cup D$ is a separation of $Y$ , choose disjoint open sets $U$ and $V$ of $X$ containing $C$ and $D$ respectively and show that $$\bigcap_{A \in \mathcal{A}} (A − (U \cup V ))$$ is nonempty.]

I started in this way:

Suppose $Y$ is not connected. Let, $C \cup D$ is a separation of $Y$ and let, $x\in C$ and $y\in D$. Since $Y\subset A$ and $A$ is connected and $X$ is Hausdorff, we can find disjoint $U_A$ (a nbh of $x$) and $V_A$ (a nbh of $y$) such that $A-(U_A\cup V_A)\neq \emptyset$. Then, $\bigcup_{x,y}[A-(U_A\cup V_A)]=C_A\neq \emptyset$. Since, $\mathcal{A}$ be a collection of simply ordered by proper inclusion, collection of $C_A$'s are too.

The confusion started when I thought that $Y$ is closed. Is that true? Can anyone help me how to proceed after what I did. Thanks.

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  • $\begingroup$ Yes, $Y$ is closed. The intersection of any family of closed sets is closed. $\endgroup$ – user194469 Sep 25 '17 at 17:10
  • $\begingroup$ @dlc thanks, even though I have no idea how to proceed. So, somehow I have to show if $\bigcap_A C_A=\emptyset$, then $Y$ can not be compact. $\endgroup$ – topology_001 Sep 25 '17 at 17:16
  • $\begingroup$ Use that the intersection of a family of closed sets with the finite intersection property in a compact space is non-empty. $\endgroup$ – Henno Brandsma Sep 25 '17 at 20:36
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You were on the right track...:

Suppose $Y$ (which is closed as an intersection of closed sets) is not connected so that we can write $Y =C \cup D$ where $C$ and $D$ are closed, non-empty and disjoint in $Y$ (and so also closed and disjoint in $X$).

So as compact Hausdorff spaces are normal we can find disjoint open sets $U$ and $V$ of $X$ such that $C \subseteq U$ and $D \subseteq V$.

Now for each $A \in \mathcal{A}$, $U \cap A \neq \emptyset$ and $V \cap A \neq \emptyset$ (as $Y \subseteq A$ etc.) So by connectedness of $A$ these $U$ and $V$ cannot cover all of $A$ (or $\{U \cap A, V \cap A\}$ disconnects $A$), so $A \setminus (U \cup V) \neq \emptyset$. Note that this set is still closed in $X$ (as a difference between a closed and an open set). It is still linearly ordered under inclusion, as the $A$ are. (we just substract a fixed set from all members of $\mathcal{A}$). So we have a family of closed sets with the finite intersection property in a compact space, so

$$\exists p \in \cap_{A \in \mathcal{A}} (A \setminus (U \cup V))$$

This $p$ is in $Y$ but not in $U$ or $V$ (so not in $C$ or $D$ either) contradiction. And so no separation of $Y$ can exist.

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  • $\begingroup$ Dear Sir , Nice proof I had only problem to show that it has finite intersection property . Please can you give me hint $\endgroup$ – SRJ Apr 3 at 13:35
  • $\begingroup$ @SRJ we have a family of non-empty sets linearly ordered by inclusion. Every finite set has a minimum. $\endgroup$ – Henno Brandsma Apr 3 at 15:41
  • $\begingroup$ Thanks a lot Sir $\endgroup$ – SRJ Apr 3 at 15:59
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Assume $C \cup D$ is a separation of $Y$. Since $C$ and $D$ are open subsets of $X$, they must be of the form $C = U \cap Y$ and $D = V \cap Y$ where $U, V$ are disjoint open subsets of $X$ (they can be chosen as disjoint since $X$ is compact Hausdorff and thus normal).

Now $Y - C \cup D = Y - U \cup V = \cap{A} - (U \cup V) = \cap(A - U \cup V) \neq \emptyset$.

Last intersection is nonempty since:

  • each of its terms is nonempty (consider the fact that neither $C$ nor $D$, and thus neither $U$ nor $V$, contains $Y$ and thus $A$; we can then use $A$'s connectedness) and of course closed

  • A set $X$ is compact if and only if every sequence of closed subsets of $X$ whose finite intersections are nonempty has a nonempty intersection (this works here due to the fact that the subsets $A$ are simply ordered).

This means that $Y - C \cup D \neq \emptyset$, i.e. we have a contradiction and $Y$ is connected.

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