Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that a function that is locally constant on a connected space is, in fact, constant. I have looked at this related question but my approach is a little different than the suggested approach and I'm unsure about the final step and would appreciate a tip. Here is what I have so far:

Let $f$ be a locally constant function on the connected space $U$. Assume that $f$ is not constant. Then, there are distinct points $x$ and $y$ such that $f(x) \neq f(y)$. Now, since $f$ is locally constant there are neighborhoods $V_x$ of $x$ and $V_y$ of $y$ such that

$$ f(V_x) = k_x, \;\; f(V_y) = k_y $$ for some constants $k_x \neq k_y$ .It follows that $V_x \cap V_y =\emptyset$

Now, let $A = U-V_x \cup V_y$ and $B = V_x \cup V_y$ so that $U = A \cup B$. With this, we have expressed $U$ as a union of disjoint sets. Since $V_x$ and $V_y$ are open $B$ is open. Note that if $A$ is empty we are done because $V_x$ and $V_y$ would comprise a separation of $U$ wich would imply that the assumption about f being not constant was faulty. So, assume $A$ is nonempty.

At this point, I want to show that $A$ itself is open. If I can do this, I believe the proof will be complete. One way I've thought about doing this is to choose a neigborhood of some point $a \in A$ and if it is not already disjoint from $B$, shrink it until it is. This would then demonstrate that $A$ is open.

Am I on the right track here?

share|improve this question
1  
Pick a point $x$. Show that the set of points $y$ such that $f(x) = f(y)$ is open (it is obviously closed). –  Qiaochu Yuan Jun 11 '11 at 22:19
2  
It is obviously closed if the codomain of $f$ is Hausdorff :) –  Mariano Suárez-Alvarez Jun 11 '11 at 22:22
    
Since you need another open set, consider the closure of $V_x$. By continuity what is the value of f on this set? –  hardmath Jun 11 '11 at 22:52
2  
@hardmath, the condition of the problem doesn't include continuity, so you'd have to prove that a locally constant function is continuous. –  Thomas Andrews Jun 11 '11 at 23:29
    
@Thomas Andrews: Point taken! That I can do... :) –  hardmath Jun 12 '11 at 1:19

5 Answers 5

up vote 7 down vote accepted

Let $\mathcal{S}$ be the set of open sets in the domain such that $f$ is constant on the open set.

Since $f$ is locally constant, we know that every $x\in \mathrm{dom}\, f$ is a member of some $S\in \mathcal{S}$.

Now, pick $x_0$, and define two sets: $U = \{x: f(x)=f(x_0)\}$ and $V=\{x: f(x)\neq f(x_0)\}$.

We can see that $U$ and $V$ are disjoint, and $U \cup V = \operatorname{dom} f$.

But each of $U$ and $V$ is just a union of open sets, namely sets in $\mathcal{S}$.

So $U$ and $V$ are both open.

share|improve this answer
    
Also, note that the notion of a "locally constant function" only depends on the topology of the domain. In fact, we don't need to know that $f$ is continuous. –  Thomas Andrews Jun 11 '11 at 23:22
    
Thanks, but I don't see how this fits into the argument I'm making above. With the way I've got everything set up is it possible to show that $A$ is open? –  ItsNotObvious Jun 11 '11 at 23:34
    
@3Sphere. My argument just shows that $A=U\setminus V_x$ is open and $V_x$ is open. So $A$ must be empty since $U$ is connected, and therefore $f$ must be constant. You don't need $y.$ –  Thomas Andrews Jun 11 '11 at 23:39

A variation and expansion of some of the ideas here:

In my first course on topology we were taught a following useful "chain-characterisation" of connectedness. Some definitions first: if $\mathcal{U}$ is a cover of a space $X$, then a chain in $\mathcal{U}$ is a finite indexed set $U_1,\ldots U_n \in \mathcal{U}$ such that for all $i=1,\ldots n-1$ we have that $U_i \cap U_{i+1} \neq \emptyset$, and it is called a chain from $x$ to $y$ in $\mathcal{U}$ (both points from $X$) when we additonally have $x \in U_1$ and $y \in U_n$.

Now, a space $X$ is connected iff for every open cover $\mathcal{U}$ of $X$ we have a chain between any pair of points of $X$.

The chain-condition implies connectedness, because if $X$ is non-connected, we have decomposition of $X$ into 2 non-empty disjoint open sets $U$ and $V$, and then for $x \in U$ and $y \in V$ there can be no chain from $x$ to $y$ in the cover $\mathcal{U} = \{U, V\}$.

The other way around is a variant of the proofs in other replies: let $\mathcal{U}$ be any open cover of $X$ and fix $x \in X$. Then define $O$ to be the set of all $y \in X$ such that there is a chain from $x$ to $y$ from $\mathcal{U}$.

$O$ is non-empty, as any $x$ in $X$ is covered by some $U \in \mathcal{U}$ and then $U_1 = U$ is a chain from $x$ to $x$, so $x$ is in $O$.

$O$ is open: let $y$ be in $O$ and let $x \in U_1,\ldots U_n$ be a witnessing chain (from $\mathcal{U}$) for it. Then for every $z$ in $U_n$, that same chain will witness that $z$ is in $O$ as well, and so $U_n \subset O$, and every point of $O$ is an interior point. Note that we do not even need the cover to be open, just that the interiors cover $X$.

$O$ is closed: suppose that $y$ is not in $O$, and let $U$ be an element from $\mathcal{U}$ that covers $y$. Suppose that some $z$ in $U$ is in $O$, and again let $x \in U_1,\ldots U_n$ be a witnessing chain for it, so with $z \in U_n$. But then the chain $x \in U_1,\ldots U_n,U_{n+1} = U$ is a chain from $\mathcal{U}$ as well, because all intersections are non-empty in the beginning by assumption, and $U_n \cap U_{n+1}$ is non-empty, as both contain $z$, and this would witness that we have a chain from $\mathcal{U}$ from $x$ to $y$. But then $y$ would be in $O$, contrary to what we assumed. So $U$ misses $O$ entirely so $O$ is closed.

But now the connectedness of $X$ forces $O = X$ (there is only one non-empty clopen set) and then we have what we wanted in the chain condition, as $x$ was arbitrary.

Having this at our disposal we are almost done: let $f$ be locally constant and for every $x$ pick a neighbourhood $U_x$ such that $f$ is constant on $U_x$. We of course consider the cover $\mathcal{U} = \{U_x : x \in X \}$, and fix $x$ in $X$. If $y$ is another point in $X$ then we have a chain from $\mathcal{U}$ from $x$ to $y$ but when 2 sets from $\mathcal{U}$ intersect, it means $f$ is constant on their union. It follows that $f(x) = f(y)$ as required.

Other applications: in a locally compact (in the sense of every point has a compact neighbourhood) connected space for every $2$ points there is a compact subset of $X$ that contains them both. Or a locally path-connected and connected space is path-connected (use path-connected open neighbourhoods, get a chain from $x$ to $y$, a glue together paths from $x$ to a point in the intersection of $U_1 \cap U_2$, a point in $U_2 \cap U_3$ etc to $y$.) and so on. It allows for all sort of local properties to get expanded more globally for connected spaces.

share|improve this answer

For each point $x \in X$, pick an open neighbourhood $U_x$ such that $f$ is constant on $U_x$. Obviously, $X = \bigcup_{x\in X} U_x$ and $U_x \neq \emptyset$ for all $x$. Now, for $x \neq y$, two things can happen:

  1. $U_x \cap U_y \neq \emptyset$ and then $U_x = U_y$, or
  2. $U_x \cap U_y = \emptyset$.

If there exist two points for which the second is true, then $X$ would not be connected. So we are in the first case for all pairs of points in $X$ and there is just one $U_x$. Hence, $f$ is constant.

share|improve this answer
2  
I don't follow. Do you mean pick a maximal open neighborhood $U_x$ such that $f$ is constant on $U_x$, or something like that? –  Qiaochu Yuan Jun 11 '11 at 23:28
    
@Qiaochu. Why maximal? –  a.r. Jun 11 '11 at 23:29
2  
Otherwise the second case is not a contradiction...? –  Qiaochu Yuan Jun 11 '11 at 23:31
    
Thanks, but how does this fit into the argument I'm making above? –  ItsNotObvious Jun 11 '11 at 23:32
    
@Quiaochu. I think it's still a contradiction -even if I haven't explained it quite right. Let's see: in the second case, take the union of all open sets $U_x$ which are different from $U_y$. Call it $U$. Then, $X = U \cup U_y$ and $X$ wouldn't be connected. Is it better now? –  a.r. Jun 11 '11 at 23:41

The fastest proof is probably the one already mentioned: pick $x$ and show that $f^{-1}(f(x))$ is clopen. Another way, which looks like Agustí's approach, is the following.

Let $f:X\to Y$ be locally constant. Define the following relation on $X$:

$x\sim y:\Leftrightarrow (f\text{ is constant on some open }U\supseteq\{x,y\}).$

It is reflexive because $f$ is locally constant. It is trivially symmetric. It is transitive because if $f$ is constant on $U$ and $V$ with $U\cap V\neq\emptyset$ then $f$ is constant on $U\cup V$. Thus $\sim$ is an equivalence relation, and we get a partition $P=\{[x_i]\ |\ i\in I\}$ of X. There is a nice description of the equivalence classes, namely

$[x]=\bigcup \{U\text{ open }\ |\ x\in U,|f(U)|=1\}$

i.e. it is the biggest open set containing $x$ on which $f$ is constant. Thus

$X=\bigcup_{i\in I}[x_i]$

is the disjoint union of non-empty opens. If $X$ is connected, then we must have $|I|=1$, i.e. $f$ is constant on $X$.

share|improve this answer

Are we don't need the hypothesis $f$ is continuous? Since if we pick $z\in X$ then $A = \{x\in X: f(x) = f(z)\}$ is open, to see this, let $y\in A$, then by $f$ is locally constant, exist open set $B_y$ containing $y$ such that $f(B_y) =\{f(y)\} =\{f(z)\}$ so $B_y \subset A$. Otherwise, $X\backslash A = \{x\in X: f(x)\neq f(z)\}$ is also open, let $w\in X\backslash A$, then exist open set $B_w$ containing $w$ such that $f(B_w) = \{f(w)\}$, so $B_w \subset X\backslash A$.\ Hence $X = A\cup (X\backslash A)$ where $A$ and $X\backslash A$ are open, by connectedness we have $X = A$ since $A\neq \emptyset$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.