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.

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    $\begingroup$ Suppose $f$ maps $x$ to the norm of $x$? (And you could make $f$ as smooth as you want, really.) $\endgroup$ – Carl Offner Jun 6 '12 at 22:16
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    $\begingroup$ Suppose $f$ maps $x$ to the norm of $x$ minus $1$. Then $f^{-1}(-\infty;0)$ is bounded, no? $\endgroup$ – Will Sawin Jun 6 '12 at 22:56
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    $\begingroup$ Take two sequences of points $p_k$ and $q_k$ going to infinity such that $f(p_k) < 0$ and $f(q_k) > 0$. Then take any curve going from $p_k$ to $q_k$ and lying outside the ball of radius $\min \{|p_k|, |q_k|\}$. Then by the intermediate value theorem, there exists a point $r_k$ on the curve so that $f(r_k) = 0$. Hence $f$ cannot be proper. $\endgroup$ – Romain Gicquaud Jun 7 '12 at 9:49
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    $\begingroup$ @Valerio : Regarding your comment about unbounded simply connected open subsets of the plane, actually we have that if $G$ is an unbounded region in the plane, then $G$ is simply connected if and only if $\partial G \cup \{\infty\}$ is connected. $\endgroup$ – Malik Younsi Jun 7 '12 at 12:59
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    $\begingroup$ It seems that the question can be reformulated as follows : Let $F$ be a closed subset of $\mathbb{R}^n$ ($n>1$) that disconnects it, with all connected components of the complement unbounded. Then $F$ has an unbounded component. $\endgroup$ – BS. Jun 23 '12 at 16:15

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 $$H_1(\mathbb{R}^n,\mathbb{R}^n-F)\simeq \widetilde{H}_0(\mathbb{R}^n-F)\neq 0\ .$$


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.


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