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I must prove the following:

Let $X_i \not= \emptyset, \ \forall i \in I$, and assume $\prod_i X_i$ is normal. Then

1) $X_i$ is $T_1$ for all $i$

2) $X_i$ is $T_4$ for all $i$

The initial product topology is the coarsest topology for which all canonical projections $\pi_i: X \mapsto X_i$ are continuous. $T_1$ means that for any two distinctive points, both have an open neighborhood that doesn't contain the other. Equivalent to that any singleton set is closed. $T_4$ is the same as normality and means that any two disjoint closed sets have disjoint open neighborhoods containing them.

Normality is not hereditary, so that cannot be used.

I considered using preimages. Because the projections are continuous, the preimage of any closed set should be closed. But I don't know what sets are closed in $X_i$.

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  • $\begingroup$ The key point here is to understand the open sets on product topology. $\endgroup$ – Sigur Feb 4 '13 at 20:48
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HINT: Let $i\in I$, and let $F$ and $G$ be disjoint closed sets in $X_i$. Then $\pi_i^{-1}[F]$ and $\pi_i^{-1}[G]$ are disjoint closed sets in $X$.

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  • $\begingroup$ So, because $X$ is normal, these two have open disjoint neighborhoods and (I checked) projection is an open map, so the projections of these two sets are also open. Are they disjoint? I'd say yes, because if they weren't, then the points in preimage of the intersection would be in both of the disjoint sets, which is impossible. Thus $X_i$ are $T_4$. Is it OK thus far? $\endgroup$ – Valtteri Feb 4 '13 at 21:21
  • $\begingroup$ @Valtteri: Yes, it is. $\endgroup$ – Brian M. Scott Feb 4 '13 at 21:34
  • $\begingroup$ @BrianMScott I still need to show that $X_i$ is $T_1$. If I have interpreted correctly, a vector $Y = (x_1,x_2,x_3...) \in X$ is a singleton if each $x_i \in X_i$ is a singleton. $X$ is normal, thus Y is closed. Thus $Y^c$ is open and its projection $\pi_i$ is open also and $\{x_i\}^c$. Thus the complement of that is closed and a singleton is closed. Sounds a bit wishful thinking to me... $\endgroup$ – Valtteri Feb 4 '13 at 22:40
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    $\begingroup$ @Valtteri: I’m assuming that your definition of normal is my definition of $T_4$, meaning normal and $T_1$. Let $x_i,y_i\in X_i$ with $x_i\ne y_i$. For $j\in I\setminus\{i\}$ fix a point $x_j=y_j\in X_j$. Then $x=\langle x_i:i\in I\rangle$ and $y=\langle y_i:i\in I\rangle$ are distinct points of $X$ that differ only in the $i$-th coordinate. A basic open set in $X$ that contains $x$ but not $y$ will easily give you an open nbhd of $x_i$ in $X_i$ that does not contain $y_i$. $\endgroup$ – Brian M. Scott Feb 5 '13 at 2:54
  • $\begingroup$ @BrianMscott Well, the terminology is a tad odd, as the professor is basically using two books in the course, the other uses one definition, the other uses another, and the other says terms $T_4$ and normal are same and both include $T_1$, but then at some times it seems to contradict that...so take your pick. But thank you very much for great hints and help. I was somewhat unsure what a "point" means in product topology, now I understand more. Much appreciated. $\endgroup$ – Valtteri Feb 5 '13 at 4:41

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