An homeomorphism in the proof : $X_i$ is connected $\implies\prod_{i\in I} X_i$ is connected.

Question

In this proof, apparently there is an "obvious homeomorphism" between $Y_F$ and $\prod X_i$.

I've thinking about it and I just get this $$f:Y_F\to\prod_{i\in F}X_i$$ with $$f(x_i)=x_i$$ so $f$ is continuous,bijective and inverse continuous, and $f(x_i)=x_i=p_i\in\prod X_{i\in F}$.

Is $f$ the correct homeomorphism ?

$\forall i\in I$, if $X_i$ is connected, then $\prod_{i\in I} X_i$ is connected.

Proof. First prove that any finite product of connected spaces is connected. This can be done by induction on the number of spaces, and for two connected spaces $X$ and $Y$ we see that $X \times Y$ is connected, by observing that $X \times Y = (\cup_{x \in X} (\{x\} \times Y)) \cup X \times \{y_0\}$, where $y_0 \in Y$ is some fixed point. This is connected because every set $\{x\} \times Y$ is homeomorphic to $Y$ (hence connected) and each of them intersects $X \times \{y_0\}$ (in $(x,y_0)$), which is also connected, as it is homeomorphic to $X$. So the union is connected by standard theorems on unions of connected sets. Now finish the induction. So for every finite set $I$, $\prod_{i \in I} X_i$ is connected.

Now if $I$ is infinite, fix points $p_i \in X_i$ for each $i$, and define $$Y = \{ (x_i) \in \prod_i X_i: \{i \in I: x_i \neq p_i \} \text{ is finite }\}$$

Now for each fixed finite subset $F \subset I$, define $Y_F = \{ (x_i) \in \prod_i X_i: \forall i \notin F: x_i = p_i \}$.

• By the obvious homeomorphism, $Y_F$ is homeomorphic to $\prod_{i \in F} X_i$, which is connected by the first paragraph.

So all $Y_F$ are connected, all contain the point $(p_i)_{i \in I}$ of $\prod_{i \in I} X_i$, and their union (over all finite subsets $F$ of $I$) equals $Y$. So again by standard theorems on the union of connected subsets of a space, $Y$ is a connected subspace of $\prod_{i \in I} X_i$.

Finally note that $Y$ is dense in $\prod_{i \in I} X_i$, because every basic open subset $O$ of the product depends on a finite subset of $I$, in the sense that $O = \prod_{i \in I} U_i$ where all $U_i \subset X_i$ are non-empty open and there is some finite subset $F \subset I$ such that $U_i = X_i$ for all $i \notin F$. Pick $q_i \in U_i$ for $i \in F$ and set $q_i = p_i $ for $i \notin F$. The $(q_i)_{i \in I}$ is in $O \cap Y_F \subset O \cap Y$, so every (basic) open subset of $\prod_{i \in I} X_i$ intersects $Y$.

Now use that the closure of a connected set is connected to conclude that $\prod_{i \in I} X_i$ is connected, also for infinite $I$.

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This proof was made by Henno here Connected topological spaces, product is connected .

Answer 1

The homeomorphism $H$ maps $x \in \prod_{i \in F} X_i$, ( so $x$ is a function defined on $F$ to $\cup_{i \in F} X_i$ such that $x(i) \in X_i$ for all $i \in F$ (we also denote $x(i)$ by $x_i$)), to $H(x) : I \to \cup_{i \in I} X_i$ defined by $H(x)(i) = x(i)$ for $i \in F$ and $H(x)(i) = p_i$ for $i \notin F$. So $x$ is a point with given "coordinates" in $X_i$ for $i \in F$ and we extend this to a point (i.e. function) on the whole product by giving it the value $p_i$ from our prechosen point for all other $i \in I$. Then by definition we get a point in $Y_F$ (as all points outside $F$ are mapped to $p_i$) and it's clearly a bijection. Continuity of $H$ is clear as $\pi_i \circ H$ (the $\pi_i$ are the standard projection maps) is either the identity for $i \in F$ or a constant map for $i \notin F$, so the universal property of the product implies $H$ is continuous from $\prod_{i \in F} X_i$ to $Y_F \subseteq \prod_{i \in I} X_i$. Its inverse is just the "projection" that maps $x \in \prod_{i \in I} X_i$ to its restriction $x|F \in \prod_{i \in F} X_i$ and which is also continuous by the same universal property applied to the finite product.

Answer 2

Well, what you've written doesn't make sense: your definition says $f(x_i)=x_i$ which would mean $x_i$ is supposed to be an element of both $Y_F$ and $\prod_{i\in F}X_i$, and I have no idea what "$f(x_i)=x_i=p_i\in\prod X_{i\in F}$" is supposed to mean. But I think what you had in mind is correct: the homeomorphism is just the projection map $Y_F\to \prod_{i\in F} X_i$ that sends $(x_i)_{i\in I}$ to $(x_i)_{i\in F}$ (i.e., we forget all the coordinates except those in $F$).