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