Connected components of the boundary of an open subset

2012-06-06T22:08:54Z

Hi!

Let f be a (continuous, $C^\infty$... whatever) function from $\mathbb{R}^n$ ($n \geq 2$) to $\mathbb{R}$. Assume that each connected component of $f^{-1} (0; \infty)$ and $f^{-1} (-\infty; 0)$ is unbounded. Also assume that $f$ changes sign so that neither $f^{-1} (0; \infty)$ nor $f^{-1} (-\infty; 0)$ is empty. Is it possible to prove that there exists at least one connected component of $f^{-1}(0)$ which is unbounded? Thanks.

Answer by BS for Connected components of the boundary of an open subset

2012-06-23T18:03:12Z

Let $F=f^{-1}(0)\subset \mathbb{R}^n$.

It is a closed subset, which disconnects $\mathbb{R}^n$, and all connected components of the complement are unbounded.

The question amounts to see if the connected component of $\infty$ in $\widehat{F}=F\cup\infty \subset S^n$ could be reduced to $\infty$.

If it were the case, there would be a decreasing sequence $V_i$ of neighbourhoods of $\infty$ with intersection only this point and $Fr(V_i)$ disjoint from $F$ (in a compact space, quasicomponents are the same as connected components).

But then $K_i=F- V_i$ is compact and does not disconnect $\mathbb{R}^n$, otherwise there would be a bounded component of the complement, since it would also disconnect $S^n$ ($n>1$ is used here). But this easily implies that $\mathbb{R}^n-F$ would have a bounded component, contrary to the assumptions.

Hence $F$ is the union of a sequence of disjoint non-disconnecting compact sets $L_i=K_i\setminus K_{i-1}$ going to infinity.

Now observe that $L_i$ doesn't disconnect any open connected subset $U$ containing it, by excision. More precisely this gives an isomorphism of relative homology groups $$H_1(U_i,U_i-L_i)\simeq H_1(\mathbb{R}^n,\mathbb{R}^n-L_i)\simeq \widetilde{H}_0(\mathbb{R}^n-L_i)=0$$, where the second isomorphism comes from the homology exact sequence of the pair $(\mathbb{R}^n,\mathbb{R}^n-L_i)$. Then the homology exact sequence for $(U_i,U_i-L_i)$ gives $\widetilde{H}_0(U_i-L_i)=0$ for a connected $U_i$, as claimed.

It remains to take disjoint connected $U_i$'s containing $L_i$'s, with union $U$ and then $$H_1(U,U-F)\simeq \bigoplus_i H_1(U_i,U_i-L_i) =0,$$ a contradiction since this is also $$H_1(\mathbb{R}^n,\mathbb{R}^n-F)\simeq \widetilde{H}_0(\mathbb{R}^n-F)\neq 0\ .$$

Answer by George Lowther for Connected components of the boundary of an open subset

2012-06-28T01:17:11Z

Yes, it is possible to prove that $f^{-1}(0)$ has an unbounded component. I'll give a proof of this. I see that BS already has one proof -- maybe that can also be translated into a similar argument. In fact, the result holds in rather more generality than that stated. We can replace $X=\mathbb{R}^n$ by any locally compact Hausdorff space which is connected, locally path connected, singly connected, and has one end. Saying that $X$ has one end means that $X\setminus C$ has a single unbounded component for each compact $C$. Note that $\mathbb{R}^n$ satisfies all of these properties for $n\ge2$, but has two ends for $n=1$ when the result fails. It also possible to generalize a bit further and, instead of requiring $X$ to be locally path connected and singly connected, replace this by the condition that the first Čech cohomology group vanishes.

The condition that $X$ is locally path connected and singly connected gives the following.

Lemma 1: If $U$ is a connected open subset of $X$ and $C\subseteq U$ is a closed set such that $U\setminus C$ is disconnected, then $X\setminus C$ is disconnected.

Proof: As $U\setminus C$ is disconnected, it can be written as the union $V \cup W$ of disjoint nonempty open sets $V,W$. Given any continuous curve $f\colon[a,b]\to X$ whose endpoints $f(a),f(b)$ are not in $C$, it is possible to define an intersection number as follows. Choose disjoint closed intervals $(s_i,t_i)$ ($i=1,2,\ldots,n$) such that $f^{-1}(C)\subseteq \bigcup_i(a_i,b_i)$ and $[a_i,b_i]\subseteq f^{-1}(U)$. Set $\epsilon_i=1$ if $f(a_i)\in V,f(b_i)\in W$. Set $\epsilon_i=-1$ if $f(a_i)\in W,f(b_i)\in V$, and $\epsilon_i=0$ otherwise. Then define the intersection number as $I(f)=\sum_i\epsilon_i$. This counts the number of times that $f$ passes through $C$, with a positive count for each time it goes from $V$ to $W$ and a negative count when it goes in the opposite direction. It can be seen that if $f$ remains within $U$ then, independently of the choice of $a_i,b_i$, we have $I(f)=1$ if $f(a)\in V,f(b)\in W$ and $I(f)=-1$ if $f(a)\in W,f(b)\in V$. Otherwise, $I(f)=0$. Similarly, if $C$ is not in the image of $f$ then $I(f)=0$. By breaking $f$ down into intervals on which its image is alternately contained in $U$ and disjoint from $C$, we see that $I(f)$ is independent of the choice of $a_i,b_i$.

Next, if $(s,t)\mapsto f_s(t)$ is a continuous map from $[0,1]$ to $X$ such that $f_s(0)$ and $f_s(1)$ are not in $C$, then it can be seen that $I(f_s)$ is independent of $s$. In fact, for a given $s_0$, choosing $a_i,b_i$ as above in the definition of $I(f_{s_0})$, it can be seen that the same choice applies for calculating $I(f_s)$ for any $s$ in a small interval about $s_0$, showing that $I(f_s)$ is locally constant and, hence constant. Therefore $I(f)=I(g)$ for curves $f,g$ which are homotopic relative to their endpoints. So, for points $P,Q\in X\setminus C$, we can define $I(P,Q)$ to be the intersection number, $I(f)$, for any curve $f$ joining $P$ to $Q$. As the space $X$ is singly connected, this is well-defined and independent of $f$.

Choose a point $P\in V$. Let $V^\prime$ be the set of points $Q\in X\setminus C$ with $I(P,Q)=0$ and $W^\prime$ be those points with $I(P,Q)\not=0$. Then, $V^\prime,W^\prime$ are disjoint open sets whose union is $X\setminus C$. If these are nonempty then it implies that $X\setminus C$ is disconnected. However, $P\in V^\prime$. Also, given any $Q\in W$ the local path connectedness implies that there is a curve joining $P$ to $Q$ in $U$, so $I(P,Q)=1$ and $Q\in W^\prime$. QED

Together with the properties that $X$ is locally compact Hausdorff and connected with one end, we can prove the result. Set $U=f^{-1}((-\infty,0))$ and $V=f^{-1}((0,\infty))$, which are disjoint open sets whose connected components are unbounded (i.e., not relatively compact). Also, set $F=X\setminus(U\cup V)$. We can show that compact connected components of $F$ only occur as isolated islands inside $U$ or $V$.

Suppose that $C$ is a compact connected component of $F$ and that $P\in C$. Then, by local compactness, there is a relatively compact open set $S$ containing $C$. From the definition of connected components, for each $Q\in\partial S$, there will be a closed subset of $F$ containing $P$ but not $Q$ and which is open in the subspace topology on $F$. Then, by compactness of $\partial S$, there is a closed subset of $F$ containing $P$ and disjoint from $\partial S$, and open in the subspace topology. Call this $K_0$. Intersecting with $S$ if necessary, we can suppose that $K_0\subseteq S$, so $K_0$ is compact. Then, by local compactness again, there is a relatively compact open set $W$ containing $K_0$ and such that $\bar W$ is disjoint from $F\setminus K_0$. In particular, $W$ is a relatively compact open set containing $P$ such that $\partial W$ is disjoint from $F$. Replacing $W$ by its connected component containing $P$, we can assume that it is connected. As $C$ is connected and intersects with $W$ but not $\partial W$, it must be contained in $W$. Suppose that $W\setminus K$ was not connected. By the lemma, $X\setminus K$ would be disconnected. Then, as $X$ has one end, $X\setminus K$ would contain at least one relatively compact connected component, say $T$. As the boundary of $T$ is contained in $W$, $T\cap W$ is nonempty. Then, $T\setminus F$ is a nonempty relatively compact connected component of $X\setminus F$, contradicting the assumptions. So, by contradiction, $W\setminus K=W\setminus F=(W\cap U)\cup(W\cap V)$ must be connected, so $W$ is disjoint from either $U$ or $V$. This implies that $\partial W$ is contained in either $U$ or $V$.

So, we have shown that every compact connected component of $F$ is contained in a relatively compact open set whose boundary is contained in either $U$ or $V$. Not all connected components of $F$ con be of this form. To show this, let $U^\prime$ be the union of all relatively compact open sets with boundary contained in $U$ and $V^\prime$ be the union of all relatively compact open sets with boundary contained in $V$. By local compactness, we have $U\subseteq U^\prime$ and $V\subseteq V^\prime$. Then, $U^\prime,V^\prime$ are disjoint open sets. Suppose not. Then, there would exist relatively compact open sets $U_0,V_0$ with boundary contained in $U,V$ respectively, and with nonempty intersection. By connectedness of $X$, there must be a point $P$ in the boundary of $U_0\cap V_0$. This will be in the boundary of either $U_0$ or $V_0$ so, wlog, take $P\in \partial U_0$. Then, $P\in U$. Also, $P\in\bar V_0$. As $P$ is not in $V$, so not on the boundary of $V_0$, this gives $P\in U\cap V_0$. So, $U\cap V_0$ is nonempty and, as its boundary is disjoint from $U$, it is closed in $U$. This means that $U\cap V_0$ contains a connected component of $U$, contradicting the conditions that $U$ has unbounded components and $V_0$ is compact.

As $X$ is connected, $X\setminus(U^\prime\cup V^\prime)$ is nonempty. Furthermore, for any $P$ in this set, the connected component of $F$ containing $P$ is not compact, by the argument above.

Connected components of the boundary of an open subset - MathOverflow

Connected components of the boundary of an open subset

Answer by BS for Connected components of the boundary of an open subset

Answer by George Lowther for Connected components of the boundary of an open subset