The main result is that a ball in a normed linear space is connected. This is the next lemma. From this, it follows that for an open set, it is connected if and only if it is arcwise connected.
Proof: This is easy from the convexity of the set. If x,y ∈ B
Proof: Let p ∈ X. Then by assumption, for any x ∈ X, there is an arc joining p and x. This arc is connected because it is the continuous image of an interval which is connected. Since x is arbitrary, every x is in a connected subset of X which contains p. Hence Cp = X and so X is connected. ■
Proof: By Proposition 3.13.2 it is only necessary to verify that if U is connected and open in the context of this theorem, then U is arcwise connected. Pick p ∈ U. Say x ∈ U satisfies P if there exists a continuous function, γ :
If x ∈ A, then Lemma 3.13.1 implies B
Then it is clear that γ1 is a continuous function mapping p to y and showing that B
Now consider B ≡ U ∖ A. I claim this is also open. If B is not open, there exists a point z ∈ B such that every open set containing z is not contained in B. Therefore, letting B
It remains to verify the connected components are open. Let z ∈ Cp where Cp is the connected component determined by p. Then picking B
As an application, consider the following corollary.
Proof: Suppose not. Then it achieves two different values, k and l≠k. Then Ω = f−1
which is the inverse image of an open set while f−1
class=”left” align=”middle”(V,W) As A Vector Space3.14. EXERCISES