0

Click to see question

Hello all,

I was working on this problem and was having trouble getting started.

My idea was to create a set, say A, for some $n \in \mathbb{N}^+$ defined $A = \{b \in F: d(z_{j-1},z_j) < \epsilon, \text{for} \space 1 \leq j \leq n \space \text{&} \space a = z_0, b = z_n\}.$ That is, A contains all $b \in F$ such that the distance between two consecutive points is less than epsilon (provided a = z0, b=zn). Then, I was going to show that A is both open and closed and since F is nonempty and connected, since it would imply that A = F. This should show that there is an $n \in \mathbb{N}^+$ and points ($z_0,...,z_n$) in F such that the conclusion is satisfied in the problem.

I wasn't sure how to show that A is open and closed. Any help to get started/the general idea would be appreciated.

Thanks.

mathmicha
  • 1
  • see the following link for a characterization of connectness using the notion of chains, as it will help you with your proof here: https://math.stackexchange.com/questions/44850/locally-constant-functions-on-connected-spaces-are-constant/44938#44938 – C Squared Sep 01 '22 at 18:19

1 Answers1

1

If you have a sequence of points from $a$ to $b$ and a point $b'$ within distance $\epsilon$ of $b$, then you can add $b'$ to the end of the sequence to obtain a sequence from $a$ to $b'$. You can use this to show $A$ is open (directly) and the complement of $A$ is open (by contradiction).

Jack Haviland
  • 21