Define for a field $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}, \mathbb{F}^\omega := \{(x_n)_{n\geq 1} : x_i \in \mathbb{F}\,\forall i\}.$ Let $x=(x_n)_{n\geq 1}, y=(y_n)_{n\geq 1} \in \mathbb{R}^\omega$. Define the $\infty$-norm or sup norm to be $d_\infty(x,y) := ||x-y||_\infty = \sup_{i\in\mathbb{N}}|x_i-y_i|.$ Define $\ell_\infty(\mathbb{C}) := \{x=(x_n)\in \mathbb{R}^\omega : ||x||_\infty < \infty\}.$ Determine whether the set
$A := \{a \in \ell_\infty(\mathbb{C}) : |a_k| = 1\,\forall k \in \mathbb{Z}^+\}$ is connected (in the space $(\ell_\infty(\mathbb{C}), d_\infty)$).
I tried several times to show that this set is connected, but I was unsuccessful each time. For instance, I thought of using the fact that a connected subset of a metric space has no nontrivial proper closed and open sets. I also thought of defining a path between any two points of $A$ but because it would map an element of $[0,1]$ to an element with infinitely many entries, I wasn’t sure how to show that such a map would be continuous.
Best Answer
Let $b=(b_k)_k=(e^{ib'_k})_k$ and $c=(c_k)_k=(e^{ic'_k})_k$ belong to $A$, with each $b'_k,c'_k\in [0,2\pi ).$
For $t\in [0,1]$ let $F(t)=(e^{i(1-t)b'_k+itc'_k})_k.$
We have $F:[0,1]\to A$ with $F(0)=b$ and $F(1)=c.$ We have $$ \|F(t)-F(t')\|=\sup_k|e^{i(1-t)b'_k+itc'_k}-e^{i(1-t')b'_k+it'c'_k}|=$$ $$=\sup_k|2\sin ((t-t')(c'_k-b'_k)/2)|\le$$ $$\le\sup_k|(t-t')(c'_k-b'_k)|\le 2\pi |t-t'|.$$ So $F$ is Lipschitz-continuous with Lipschitz constant $2\pi.$ So $F$ is a path in $A$ from $b$ to $c.$ So $A$ is path-connected, so $A$ is connected.