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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...

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

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