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mahler1
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Homework Statement .
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?
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?