I am struggling on solving the problem related to the connectedness of a topological space:
Let $X$ be a countable product of a closed interval $[0, 1] \subset \mathbb{R}$, given the product topology. Prove or disprove that the space $ Z := X \setminus \{ ( 1, 1, 1, \cdots ) \}$ is connected.
I tried to solved this problem by showing that $Z$ is actually path-connected:
Since the space $Z' := [0, 1]^2 \setminus \{(1, 1)\}$ is path-connected, for every point $x, y \in Z'$, there is a path $p =(p_1, p_2) : [0, 1] \to Z'$ such that $p(0) = x$ and $p(1)=y$. Now take two points ${\bf x} = (x_n), {\bf y} = (y_n) \in Z$. Consider the function $q : [0, 1] \to Z$ defined by $$ q(t) = (p_1(t), p_2(t), p_3(t), \ldots ) $$ where the function $t \mapsto (p_1(t), p_2(t))$ is a path from $(x_1, x_2)$ to $(y_1, y_2)$ in $Z'$, and $p_j$ denotes the line segment from $x_j$ to $y_j$ for $j \geq 3$. Since each coordinate function of $q$ is continuous, so is $q$ since $Z$ is given the product topology. Hence $Z$ is path-connected, which implies the connectedness of $Z$.
My questions are,
(1) Is my trial correct? It looks like there is some ambiguous point in using the path in $Z'$;
(2) Since the path-connectedness is more stronger property than the connectedness, I wonder whether there is an approach which is related only to the connectedness property, not the path-connectedness.