# Homework Help: Connected components of a metric subspace

1. Oct 29, 2013

### mahler1

The problem statement, all variables and given/known data.
Consider the subspace $U$ of the metric space $(C[0,1],d_∞)$ defined as $U=\{f \in C[0,1] : f(x)≠0 \forall x \in [0,1] \}$. Prove that $U$ is open and find its connected components.

The attempt at a solution.
First I've proved that $U$ is open. I want to check if my proof is correct:
Let $f \in U$. $f(x)≠0$ and $f$ is continuous, this means $f(x)>0$ or $f(x)<0$. Suppose $f(x)>0$. The function $f$ is a continuous function defined on a compact set then there exists $x_0 \in [0,1]$ such that $f(x)≥f(x_0)>0 \forall x \in [0,1]$. Consider the ball $B(f,\frac{f(x_0)}{2})$. Let $g \in B(f,\frac{f(x_0)}{2})$. Then, $|f(x)-g(x)|≤sup_{x \in [0,1]}|f(x)-g(x)|=d_∞(f(x),g(x))<\frac{f(x_0)}{2}$.But then $f(x)-g(x)<\frac{f(x_0)}{2}→0<f(x)-\frac{f(x_0)}{2}<g(x)$. This proves that $g \in U$ for an arbitrary $g \in B(f,\frac{f(x_0)}{2})→B(f,\frac{f(x_0)}{2}) \subset U$

The case $f(x)<0$ is analogue. $f$ is a continuous function on a compact set, then there exists $x_1 \in [0,1]$ such that $f(x)≤f(x_1)<0 \forall x \in [0,1]$. Consider the ball $B(f,\frac{-f(x_1)}{2})$. Let $g \in B(f,\frac{-f(x_1)}{2})$. Then $|f(x)-g(x)|≤sup_{x \in [0,1]}|f(x)-g(x)|=d_∞(f(x),g(x))<\frac{-f(x_1)}{2}$. This means $\frac{f(x_1)}{2}<f(x)-g(x)→0<\frac{f(x_1)}{2}-f(x)<-g(x)→0>f(x)-\frac{f(x_1)}{2}>g(x)$. This implies $g \in U$ for an arbitrary $g \in B(f,\frac{f(x_1)}{2})→B(f,\frac{f(x_1)}{2}) \subset U$

I am stuck on the second part of the problem. If a function f is in the subspace U, then the connected component of f would be the union of all the connected subspaces in U that contain f. My doubt is: What are the connected components of a subspace (in this case, U)?. How could I try to find them?

2. Oct 29, 2013

### Dick

You've already got the ingredients you need. Define $U_{+}$ to be the set of positive functions and $U_{-}$ to be the set of negative functions. You've proved both of those are open. So? BTW what definition of connectedness are you using?

3. Oct 29, 2013

### pasmith

$U$ can be written as a disjoint union of open sets, namely
$$U = \{ f \in U: f(x) > 0\} \cup \{f \in U : f(x) < 0\}$$

It is possible to show that each of those subsets is path connected, and hence connected.

4. Oct 29, 2013

### mahler1

If $(X,d)$ is a metric space and $C \subset X$, then $C$ is connected if:
1) There aren't any $U$,$V$ non-empty, disjoint and open in $C$ such that $C=U \cup V$
2) If $A \subset C$ is non-empty, open and closed in $C$, then $A=C$.

These are the two equivalent definitions I am using.

So, you've said that I should consider $U_{+}$ and $U_{-}$. I've already proved they are open and it is easy to see that both are non-empty ($f(x)$ constantly $1$ and $h(x)$ constantly $-1$ are in $U_{+}$ and $U_{-}$ respectively) and it's immediate to show that they are disjoint for if $f \in U_{+} \cap U_{-}$ then $0<f(x)<0 \forall x \in [0,1]$, which is absurd. From here I can conclude that $U$ is not connected. I am pretty sure that $U_{+}$ and $U_{-}$ are the only connected components of $U$, but I don't know how to prove it. In fact I am not sure of the formal definition of connected components of a subspace. I intuitively think that they are the "maximal" connected subsets of the space (in terms of inclusion). Is that the definition?

5. Oct 29, 2013

### mahler1

Ok, once I've proved that, should I prove that if $A \subset U$ is connected then $A \subset U_{+}$ or $A \subset U_{-}$? (using Dick's notation)

6. Oct 29, 2013

### Dick

Right you are. Now you just need to show $U_{+}$ and $U_{-}$ are connected.

7. Oct 29, 2013

### Dick

One way is to show that they arcwise connected. If $f_1$ and $f_2$ are two functions in $U_{+}$, then $t f_1+(1-t) f_2$ is in $U_{+}$ for t in [0,1], right? Suppose that there were two sets that disconnect $U_{+}$ but $f_1$ is in one and $f_2$ is in the other?

Last edited: Oct 29, 2013
8. Oct 29, 2013

### mahler1

I've just proved that $U_{+}$ and $U_{-}$ are path connected by the same way you did.

Sorry, I don't know if this was what you were pointing me out but it is what I've came up with:

Suppose $A \subset U$ is connected and suppose $A$ is not totally included in $U_{+}$ or
$U_{-}$. We know that $U_{+}$ and $U_{-}$ disconnect $U$, so $A \subset U_{+} \cup U_{-}$. By the previous statements we have $A \cap U_{+}≠∅$ and $A \cap U_{-}≠ ∅$. I define $A_{+}=\{f \in A : f \in U_{+}\}$ and $A_{-}=\{h \in A : h \in U_{-}\}$. Then $A_{+}$ and $A_{-}$ disconnect $A$ which is absurd since $A$ is connected. It follows $A \subset U_{+}$ or $A \subset U_{-}$. This proves $U_{+}$ and $U_{-}$ are the only connected components of $U$.

Is this correct? Thanks for your help!

9. Oct 29, 2013

### Dick

It's all fine up until "This proves $U_{+}$ and $U_{-}$ are the only connected components of $U$". It doesn't prove that. Suppose $U_{+}$ could be split into two connected components? The point to what I was telling you is that if $U_{+}$ could be disconnected then the real interval [0,1] could be disconnected. And we know it can't. That's what the arcwise connected bit is about. Maybe there is a another way to do this, but this is the way I usually think about it.

Last edited: Oct 29, 2013