Prove that a locally constant function is constant on a connected X

[docnet]

Homework Statement

Prove that a locally constant function is constant on a connected topological space X.

Relevant Equations

:)

Let $X$ be a topological space and $Y$ a set. A function $f: X \to Y$ is said to be locally constant if, for every $x \in X$, there is an open set $U$ containing $x$ so that the restriction $f|_U: U \to Y$ is constant. Prove that if $X$ is connected, a locally constant function on $X$ is constant.

Proof: $X$ is a connected topological space and $f:X\rightarrow Y$ is a locally constant function from $X$ to a set $Y$. A function is locally constant iff $\forall x_0\in X,$ there exists a neighborhood $U$ of $x$ so that $\forall x\in U, f(x)=f(x_0)$. Suppose there exists nonempty open subsets $U$ and $V$ in $X$ so that for all $x\in U$ and $y\in V$ the function restricted to $U$ and $V$ take $x$ and $y$ into different elements in $Y$, so that $f|_U(u)=u'$ and $f|_V(v)=v'$ and $u'\neq v$. If $U\cap V\neq \emptyset,$ then the function is not well defined over $U\cap V$ which leads to a contradiction to the assumption that $f$ is defined $\forall x\in X$. If $U\cap V= \emptyset$, it implies that $X$ is disconnected, which leads me to a contradiction. $\therefore f$ is constant for any $x\in X$.

Is my proof correct, and if no, how could I improve it?

 

[andrewkirk]

If $U\cap V= \emptyset$, it implies that $X$ is disconnected

I'm afraid that doesn't follow. U and V could be disjoint but have a third set that intersects them both. To be disconnected, we must demonstrate a partition of the space into disjoint open sets. It's not enough to just find two disjoint open sets.

Consider the set $C = \{f^{-1}(y) |\ y\in Y\}$. Are all the elements of C open sets? Does C form a partition of $X$?

 

[docnet]

I'm afraid that doesn't follow. U and V could be disjoint but have a third set that intersects them both. To be disconnected, we must demonstrate a partition of the space into disjoint open sets. It's not enough to just find two disjoint open sets.

Consider the set $C = \{f^{-1}(y) |\ y\in Y\}$. Are all the elements of C open sets? Does C form a partition of $X$?

I see what you mean! Do the elements of C form a partition of $X$ if $f^{-1}$ is bijective and Y itself is a topological space that it is partitioned into disjoint open sets?

 

[docnet]

If X is connected, then C is not partitioned into disjoint open sets and f will take every $x\in X$ into one constant element of Y. Can I argue that the largest open neighborhood of x is X, since X is connected, so that f is constant in all of X? I am not sure whether I can assume the openess of Y, the continuousity of f, or even that Y is a topological space from what is given!

 

[andrewkirk]

I am not sure whether I can assume the openess of Y, the continuoty of f, or even that Y is a topological space from the problem!

We don't need f to be continuous, or Y to be a topological space. Nor do we need f to be bijective, as per your earlier post. As for the 'openness of Y', that means nothing if Y is not topological. However I think you may have meant the openness of $f^{-1}(y)$ for any $y\in Y$. We need to prove that. It is fairly straightforward, given that f is locally constant. If you have covered the notion of 'interior points' you can use that fact that local constancy implies that every element of $f^{-1}(y)$ is an interior point, which is equivalent to $f^{-1}(y)$ being open. If you haven't covered that notion, you may need to do a bit more work, for instance showing that $f^{-1}(y)$ can be expressed as a union of the open neighbourhoods, and that unions of open sets are always open, even if infinitely many.

 

[docnet]

We don't need f to be continuous, or Y to be a topological space. Nor do we need f to be bijective, as per your earlier post. As for the 'openness of Y', that means nothing if Y is not topological. However I think you may have meant the openness of $f^{-1}(y)$ for any $y\in Y$. We need to prove that. It is fairly straightforward, given that f is locally constant. If you have covered the notion of 'interior points' you can use that fact that local constancy implies that every element of $f^{-1}(y)$ is an interior point, which is equivalent to $f^{-1}(y)$ being open. If you haven't covered that notion, you may need to do a bit more work, for instance showing that $f^{-1}(y)$ can be expressed as a union of the open neighbourhoods, and that unions of open sets are always open, even if infinitely many.

Sorry for sounding clueluess! what is the logical utility of proving that $f^{-1}(y)$ is open, because isn't it the same as $U \subset X$ which is given to be open? ?

 

[PeroK]

Homework Statement:: Prove that a locally constant function is constant on a connected topological space X.
Let $X$ be a topological space and $Y$ a set. A function $f: X \to Y$ is said to be locally constant if, for every $x \in X$, there is an open set $U$ containing $x$ so that the restriction $f|_U: U \to Y$ is constant. Prove that if $X$ is connected, a locally constant function on $X$ is constant.

Proof: $X$ is a connected topological space and $f:X\rightarrow Y$ is a locally constant function from $X$ to a set $Y$. A function is locally constant iff $\forall x_0\in X,$ there exists a neighborhood $U$ of $x$ so that $\forall x\in U, f(x)=f(x_0)$. Suppose there exists nonempty open subsets $U$ and $V$ in $X$ so that for all $x\in U$ and $y\in V$ the function restricted to $U$ and $V$ take $x$ and $y$ into different elements in $Y$, so that $f|_U(u)=u'$ and $f|_V(v)=v'$ and $u'\neq v$. If $U\cap V\neq \emptyset,$ then the function is not well defined over $U\cap V$ which leads to a contradiction to the assumption that $f$ is defined $\forall x\in X$. If $U\cap V= \emptyset$, it implies that $X$ is disconnected, which leads me to a contradiction. $\therefore f$ is constant for any $x\in X$.

Is my proof correct, and if no, how could I improve it?

You seem to have jumped into analytic topology at quite an advanced level without having done much proof writing. Note that it is always trivial to prove anything mathematically - even the Riemann hypothesis can be proved very simply in many ways. However, all these proofs would be wrong! Anyone can bash out a proof like you did above if they have the mindset of getting to the final answer in as few steps as possible.
You need to completely change your mindset to self-analyse every step in your proof - instead of being uncritical of your own proofs you need to challenge yourself. Especially, looking for counterexamples at every step. At the moment you are using us to home in on a proof by a process of trial and error.

This is where it's fundamentally different from "computational" mathematics. If you do a whole bunch of difficult calculations and get the book answer, then the chances are your working is correct. But, proofs are not like that. Just because you get the right answer, doesn't mean even a single step in your proof is valid. You should think about that.

In this case, I would try a simpler problem first but one that is highly relevant:

Proof that $X$ is disconnected iff there exists a continuous function $f: X \rightarrow \{0, 1\}$, where $f$ is onto (i.e. takes both values $0$ and $1$).

That, in fact, is the alternative definition of a disconnected set. In any case, that should give you a clue as to how to tackle this problem.

 

[docnet]

You seem to have jumped into analytic topology at quite an advanced level without having done much proof writing. Note that it is always trivial to prove anything mathematically - even the Riemann hypothesis can be proved very simply in many ways. However, all these proofs would be wrong! Anyone can bash out a proof like you did above if they have the mindset of getting to the final answer in as few steps as possible.
You need to completely change your mindset to self-analyse every step in your proof - instead of being uncritical of your own proofs you need to challenge yourself. Especially, looking for counterexamples at every step. At the moment you are using us to home in on a proof by a process of trial and error.

This is where it's fundamentally different from "computational" mathematics. If you do a whole bunch of difficult calculations and get the book answer, then the chances are your working is correct. But, proofs are not like that. Just because you get the right answer, doesn't mean even a single step in your proof is valid. You should think about that.

In this case, I would try a simpler problem first but one that is highly relevant:

Proof that $X$ is disconnected iff there exists a continuous function $f: X \rightarrow \{0, 1\}$, where $f$ is onto (i.e. takes both values $0$ and $1$).

That, in fact, is the alternative definition of a disconnected set. In any case, that should give you a clue as to how to tackle this problem.

Thank you for saying this! I am taking this to heart. It sounds like you're saying that writing proofs is unfamiliar territory and it is difficult for beginners, so I need to work at it and maybe eventually I will become good at it with practice. Honesty is as good as it gets, really!

To try the simpler problem:

$\{0, 1\}$ is disconnected, and is a union of the disjoint open sets $\{0\}$ and $\{1\}$. Because f is continuous, $f^{-1}(0)$and $f^{-1}(1)$ are open in X. And, because f is surjective $f^{-1}(0)\cup f^{-1}(1)= f^{-1}(0\cup 1)=X$. To check whether X is connected or disconnected, I write the intersection of the preimages. i.e. $f^{-1}(0)\cap f^{-1}(1)=f^{-1}(0\cap 1)=f^{-1}(\emptyset)=\emptyset$ which implies that X is disconnected.

 

[PeroK]

$\{0, 1\}$ is disconnected, and is a union of the disjoint open sets $\{0\}$ and $\{1\}$. Because f is continuous, $f^{-1}(0)$and $f^{-1}(1)$ are open in X. And, because f is surjective $f^{-1}(0)\cup f^{-1}(1)= f^{-1}(0\cup 1)=X$. To check whether X is connected or disconnected, I write the intersection of the preimages. i.e. $f^{-1}(0)\cap f^{-1}(1)=f^{-1}(0\cap 1)=f^{-1}(\emptyset)=\emptyset$ which implies that X is disconnected.

This is nearly right for the case that the existence of such a function implies disconnectedness. The key points are:

$f^{-1}(0)$ and $f^{-1}(1)$ are non-empty, open, disjoint and cover $X$ (i.e. their union equals $X$).

I would have put that up front of the proof as "we need to show ...". That's four things you need to prove, then you are done.

You also need the reverse implication that for any disconnected set you can find such a function.

 

[docnet]

You also need the reverse implication that for any disconnected set you can find such a function.

Did you mean I should explain that $f:X\rightarrow Y$ is surjective so the preimages of disjoint, nonempty, and open subsets in Y are taken into disjoint, nonempty, and open subsets in X. And, because f is continuous, the union of the preimage of the partitions of Y is the preimage of the union of the partitions that cover Y, thus $f^{-1}(Y)$ covers all of X?

 

[PeroK]

Did you mean I should explain that $f:X\rightarrow Y$ is surjective so the preimages of disjoint, nonempty, and open subsets in Y are taken into disjoint, nonempty, and open subsets in X. And, because f is continuous, the union of the preimage of the partitions of Y is the preimage of the union of the partitions that cover Y, thus $f^{-1}(Y)$ covers all of X?

Yes, try to justify each by a simple statement. It seems important to me to recognise these four conditions and show that each is met.

Then try the converse.

 

[PeroK]

For example, it's the fact that $f$ is onto that ensures the preimages are non-empty.

You don't need to go to town proving that as it's fairly obvious. But recognising it is important.

 

[mathwonk]

If f is not constant and takes value c somewhere, it seems that the set A where f equals c and the set B where it does not, are disjoint, non empty, and both open. Done.

 

[docnet]

In this case, I would try a simpler problem first but one that is highly relevant:

Proof that $X$ is disconnected iff there exists a continuous function $f: X \rightarrow \{0, 1\}$, where $f$ is onto (i.e. takes both values $0$ and $1$).

$f$ is onto, which means it takes both values 0 and 1.

0 and 1 are disjoint elements in the discrete topology. $\{0\}\cap \{1\}=\emptyset$

f is continuous, so the preimage of the intersection is the intersection of the preimages, i.e., $f^{-1}(\{0\})\cap f^{-1}( \{1\})=f^{-1}(\{0\}\cap \{1\})=\emptyset$.

So $X$ is a disjoint union of two open sets (open because $\{0\}$ and $\{1\}$ are open in the discrete topology, and $f$ is a continuous map)..

 

[PeroK]

$f$ is onto, which means it takes both values 0 and 1.

0 and 1 are disjoint elements in the discrete topology. $\{0\}\cap \{1\}=\emptyset$

f is continuous, so the preimage of the intersection is the intersection of the preimages, i.e., $f^{-1}(\{0\})\cap f^{-1}( \{1\})=f^{-1}(\{0\}\cap \{1\})=\emptyset$.

So $X$ is a disjoint union of two open sets (open because $\{0\}$ and $\{1\}$ are open in the discrete topology, and $f$ is a continuous map)..

That's doesn't look like a proof. Any proof must look something like:

Let $X$ be disconnected ... define $f$ as it relates to $X$ ... show $f$ has the stated properties.

Let $f$ have the stated properties ... show that $X$ is disconnected.

In other words, we must show the two-way implication "iff".

 

[docnet]

That's doesn't look like a proof. Any proof must look something like:

Let $X$ be disconnected ... define $f$ as it relates to $X$ ... show $f$ has the stated properties.

Let $f$ have the stated properties ... show that $X$ is disconnected.

In other words, we must show the two-way implication "iff".

00ps sorry

Prove that $X$ is disconnected iff there exists a continuous function $f:X\to \{0,1\}$ where $f$ is onto.

The existence of a continuous and onto $f:X\to \{0,1\}$ ensures that $X$ is disconnected. Assume that such a function exists. $f$ being onto ensures that it takes both values 0 and 1. $f$ being continuous ensures that $f^{-1}(\{0\})\cap f^{-1}( \{1\})=f^{-1}(\{0\}\cap \{1\})=\emptyset$. So $X$ is a disjoint union of $f^{-1}(\{0\})$ and $f^{-1}( \{1\})$ and thus X is disconnected.

On the other hand, $X$ being disconnected ensures the existence of such a function. Because $X$ is disconnected, there exist disjoint open sets $U$ and $V$ so that $U\cup V=X$. Define a function $$f=\begin{cases} 0 \quad \text{for}\quad U\\ 1\quad \text{for}\quad V \end{cases}$$ and we are done.

 

[PeroK]

00ps sorry

Prove that $X$ is disconnected iff there exists a continuous function $f:X\to \{0,1\}$ where $f$ is onto.

The existence of a continuous and onto $f:X\to \{0,1\}$ ensures that $X$ is connected. Assume there exists a function $f:x\to\{0,1\}$ which is onto. $f$ being onto ensures that it takes both values 0 and 1. $f$ being continuous ensures that $f^{-1}(\{0\})\cap f^{-1}( \{1\})=f^{-1}(\{0\}\cap \{1\})=\emptyset$. So $X$ is a disjoint union of $f^{-1}(\{0\})$ and $f^{-1}( \{1\})$ and thus X is disconnected.

On the other hand, $X$ being disconnected ensures the existence of a continuous function $f:X\to \{0,1\}$ where $f$ is onto. Because $X$ is disconnected there exist disjoint open sets $U$ and $V$ s.t. $U\cup V=X$. Define a function $f$ such that $f(U)=0$ and $f(V)=1$ and we are done.

That's okay, but short on precise justification. This is one reason more difficult proofs go wrong. You should use these simple proofs as training for the harder cases.

 

[docnet]

Thank you @PeroK

trying again...

Prove that $X$ is disconnected iff there exists a continuous function $f:X\to \{0,1\}$ where $f$ is onto.

The existence of a continuous and onto $f:X\to \{0,1\}$ ensures that $X$ is disconnected. Assume that such a function exists. $f$ being onto ensures that the image of $X$ covers the entirety of $\{0,1\}$. so $f$ takes both values 0 and 1. The function is continuous on all of X implies it is defined on all of X, which ensures that the preimage of $\{0,1\}$ covers the entirety of $X$. $f$ being continuous ensures that the preimages of intersections are the intersections of preimages. In other words, the preimage of disjoint sets is disjoint. 0 and 1 are disjoint sets, so their preimages are also disjoint and since they cover all of X, X is disconnected.

On the other hand, $X$ being disconnected ensures the existence of such a function. Because $X$ is disconnected, there exist disjoint open sets $U$ and $V$ so that $U\cup V=X$. Define a function $$f=\begin{cases} 0 \quad \text{for}\quad U\\ 1\quad \text{for}\quad V \end{cases}$$ and we are done.

 

[PeroK]

What I had more in mind is this:

Assume $X$ is disconnected. Then $X = A \cup B$ where:

1) $A, B$ are non-empty.

2) $A, B$ are open.

3) $A, B$ are disjoint (i.e. $A \cap B = \emptyset$)

We define $f: X \rightarrow \{0, 1\}$ by
$$f(x)=\begin{cases} 0 \ \ \text{if} \ \ x \in A \\ 1 \ \ \text{if} \ \ x \in B \end{cases}$$ Now, this is what I mean by justification:

$f$ is a well-defined function because $A \cap B = \emptyset$ [Did you see that?]

$f$ is onto because of $A, B$ are non-empty.

$f$ is continuous because the pre-image of every open set in $\{0, 1\}$ is open. In particular $f^{-1}(\{0\}) = A$ is open and $f^{-1}(\{1\}) = B$ is open.

Now, assume that such an $f$ exists:

a) $f: X \rightarrow \{0, 1\}$

b) $f$ is onto.

c) $f$ is continuous.

Let $A = f^{-1}(\{0\})$ and $B = f^{-1}(\{1\})$.

As $f$ is onto, $\exists x, y \in X$ such that $f(x) = 0$ and $f(y) = 1$. Therefore, $x \in A$ and $y \in B$, hence 1) holds (i.e. $A, B$ are non-empty).

As $f$ is continuous, $A, B$ are open (as $\{0\}$ and $\{1\}$ are open in $\{0, 1\}$).
Hence, 2) holds.

$X = A \cup B$ (We could assume this is obvious, or we could do:

Let $x \in X$. Either $f(x) = 0$ or $f(x) = 1$. Therefore, either $x \in A$ or $x \in B$, hence $x \in A \cup B$ and, as $x$ was arbitrary, $X = A \cup B$).

Hence 3) holds. And we see that $X$ is disconnected.

 

[PeroK]

The point about that proof is not that it contains every last detail, but that it mentions every apsect of what we are trying to prove. We wanted to show that:

$X = A \cup B$ where:

1) $A, B$ are non-empty.

2) $A, B$ are open.

3) $A, B$ are disjoint (i.e. $A \cap B = \emptyset$)

Is equivalent to:

There exists a function $f$ where:

a) $f: X \rightarrow \{0, 1\}$

b) $f$ is onto.

c) $f$ is continuous.

In doing this, I explicitly used all three properties 1), 2) and 3) for the "if" implication. And, I explicitly used all three properties a), b) and c) for the "only if" implication.

For pure mathematics, that is critical in terms of justififying the equivalence between the two sets of properties.

 

[docnet]

@PeroK thank you! you just taught me that iff statements need to be proven throughly in forward and backward directions. I did not know that. I feel like I should be paying you