# Queris related to “An arbitrary product $X=\prod_{\alpha \in J} X_\alpha$ is connected iff $X_{\alpha }$ is connected for each $\alpha\in J$.”

The following proof of

## An arbitrary product $$X=\prod_{\alpha \in J} X_\alpha$$ is connected iff $$X_{\alpha }$$ is connected for each $$\alpha\in J$$ is provided by

my professor.I've some queries in this proof.

Let X is connected then we shall show that each $$X_{\alpha}$$ is connected,where $$\alpha\in J,$$some index set.

Consider $$\pi_{\alpha}:X\rightarrow X_{\alpha},{\alpha}\in J$$.

Since $$\pi_{\alpha}$$ is a projection map so it is continuous & $$X$$ is connected by assumption.

Using the result:"Continuous image of a connected set is connected."

we have $$X_{\alpha}$$ is connected.

## Conversely,

Result Used:

X is connected iff every continuous function $$f:X\rightarrow$${$$0,1$$} is constant.

Fix $$f:X\rightarrow$${$$0,1$$} a continuous function.

Fix $$(a_{\alpha})\in X$$ and for each $$\alpha\in J,$$define $$\varphi_{\alpha}:X_{\alpha}\rightarrow X$$ by $$\varphi_{\alpha}o\pi_{\beta}(x) = \begin{cases} a_{\beta}, & \text{if \beta\neq \alpha} \\[2ex] x, & \text{if \beta=\alpha } \end{cases}$$.

Let $$g_{\alpha}=fo\varphi_{\alpha}:X_{\alpha}\rightarrow$${$$0,1$$} for each $$\alpha\in J.$$Then each $$g_{\alpha}$$ is constant since $$X$$ is connected.

Hence,for each $$\alpha\in J, f$$ is constant on $$\varphi [X_{\alpha}].$$

Say,Without loss of generality,for some $$\alpha_0\in J,f[\varphi_{\alpha_0[X_{\alpha}]}]=$${$$0$$}.Since,$$(a_{\alpha})\in \varphi_{\beta}[X_{\beta}]$$ for each $$\beta \in J,$$ it must be that $$\forall \alpha\in J,f[\varphi_{\alpha_0[X_{\alpha}]}]=$${$$0$$}.

Then $$f^{-1}$$[{$$0$$}] is an open subset of $$X$$ by continuity which contains all of these elements.$$\tag{*}$$

## 1.(Please clarify how $$f^{-1}$$[{$$0$$}] is open subset of $$X$$ :Is it because {0} is open in {0,1},$$f$$ is continuous,so inverse image of open set is open? )

Let $$U\subset f^{-1}$$[{$$0$$}] be a basis element of $$X$$ containing $$(a_{\alpha})$$ so that $$\pi_{\alpha}[U]=X_{\alpha}$$ for all but finitely many $$\alpha \in J$$.Let K be the finite subset of $$J$$ comprised of these $$\alpha$$ and let $$(x_{\alpha})\in U$$ be that element such that $$x_{\alpha}=a_{\alpha}$$ if $$\alpha \in K.$$

Changing the coordinates of $$(x_{\alpha})$$ indexed by $$K$$ one at a time,it follows from ($$*$$) that each time we change it,it remains in $$f^{-1}$$[{$$0$$}].Since $$\pi_{\alpha}[U]=X_{\alpha}$$ for all but finitely many $$\alpha \in J$$.

We have shown that $$\prod_{\alpha \in J} X_\alpha=f^{-1}$$[{$$0$$}].(##2.I'm not getting this sudden transition.Please explain!!)

If there is some scope of refinement in the above proof,feel free to express your views.Also check this proof critically...

• In response to 1: yes – mathworker21 Dec 12 '18 at 8:59
• The definition of $\varphi$ is missing a subscript. – William Elliot Dec 13 '18 at 2:38
• @William Elliot:see the edit – P.Styles Dec 13 '18 at 3:31
• $\pi and \varphi$ appear to be in reverse order. – William Elliot Dec 13 '18 at 7:49
• 2. The leap of faith likely depends upon showing finite products of connected sets are connected. – William Elliot Dec 13 '18 at 7:51