# Locally Constant Functions on Connected Spaces are Constant

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?

• Pick a point $x$. Show that the set of points $y$ such that $f(x) = f(y)$ is open (it is obviously closed). Jun 11, 2011 at 22:19
• It is obviously closed if the codomain of $f$ is Hausdorff :) Jun 11, 2011 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? Jun 11, 2011 at 22:52
• @hardmath, the condition of the problem doesn't include continuity, so you'd have to prove that a locally constant function is continuous. Jun 11, 2011 at 23:29
• @Thomas Andrews: Point taken! That I can do... :) Jun 12, 2011 at 1:19

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.

• 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. Jun 11, 2011 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? Jun 11, 2011 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 $y.$ Jun 11, 2011 at 23:39
• @ThomasAndrews You say one does not use continuity here, but how can I show that the locally constant function is continuous? Mar 8, 2017 at 13:57
• Letting $U_a=\{x:f(x)=a\}=f^{-1}(\{a\})$ for $a\in Y$, we see that $f^{-1}(V)=\union_{a\in V} U_a$. So $f^{-1}(V)$ is open for any subset of $Y$, and thus, in particular, or any open subset of $Y$. @user123 Mar 8, 2017 at 14:45

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 in $$\mathcal{U}$$ 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 reverse implication 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.

• This is cool! Is there a book that discusses this characterization? Feb 10, 2018 at 0:34
• @LucasSilva not that I know of. It was in our lecture notes. The idea occurs in many proofs as a proof technique, but not as a characterisation that is treated separately. Feb 10, 2018 at 6:14
• @LucasSilva it's exercise 6.3.1 in Engelking's general topology. It's unattributed (so "folklore" probably). May 21, 2018 at 22:24

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 $A\neq \emptyset$

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.

• 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? Jun 11, 2011 at 23:28
• @Qiaochu. Why maximal? Jun 11, 2011 at 23:29
• Otherwise the second case is not a contradiction...? Jun 11, 2011 at 23:31
• Thanks, but how does this fit into the argument I'm making above? Jun 11, 2011 at 23:32
• @Augusti: your argument certainly can be polished into a convincing proof. One way: for each $x \in X$ fix a neighborhood $U_x$ on which $f$ is constant (not necessarily maximal, just containing $x$). Since $f$ has at least one value, choose one value and call it $y$. Let $U = \bigcup \{ U_x : f(x) = y\}$ and $V = \bigcup \{ U_x : f(x) \not = y\}$. Then $X = U \cup V$ and $U,V$ are open and disjoint. Since $U$ is nonempty and $X$ is connected, $V$ is empty, so $f$ takes only the value $y$. Jun 12, 2011 at 3:14

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 $X$.

I think there another argument that hasn't been mentioned here yet. Let $$f:X\to S$$ be a function from a topological space $$X$$ to a set $$S$$. Then $$f$$ is continuous when $$S$$ is endowed with the discrete topology if and only if for all $$x\in X$$ there is a neighborhood $$U\subset X$$ of $$x$$ such that $$f|_U$$ is constant. On this case, we say that $$f$$ is locally constant. In particular, if $$X$$ is also connected, then $$f(X)$$ is connected since continuity preserves connectedness. Since $$S$$ is totally disconnected (as discrete spaces are), $$f(X)$$ is a point.