Let $X = \prod_i X_i$ and let $v \in X$. Define $V = \{x \in X : x_i = v_i \text{ for $i \ne j$} \}$. Show that $V$ is homeomorphic to $X_j$.

Question

Let $X = \prod_i X_i$ and let $v \in X$. Define $V = \{x \in X : x_i = v_i \text{ for $i \ne j$} \}$, where both $i$ and $j$ belong to some indexing set. Show that $V$ is homeomorphic to $X_j$.

I am bit confused on what is the set $V$ here. It seems that it's the set of all $x$ in the product space such that the $i$'th component equals the $i$'th component of some $v \in X$ for $i \ne j$. Why is there a condition $i \ne j$?

Also to show that these spaces are homeomorphic if I define $f: V \to X_j$ I need to construct this to be continuous bijection with continuous inverse. However I think have a problem trying to construct this since I don't really understand what $V$ is. Any hints on what to do?

Answer 1

I think $j$ is suposed to be fixed.So $V$ consists of all points whose $j-$th coordinate is arbitarry but all other coordinates are those of a fixed element $v$. The map you require is $f(x)=x_j$. Can you check that this is a homeomorphism?