Hausdorffness of the product topology

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quasar987
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Main Question or Discussion Point

[SOLVED] Hausdorffness of the product topology

Is it me, or is the product of an infinite number of Hausdorff spaces never Hausdorff???

Recall that the product topology on

[tex]\Pi_{i\in I}X_i[/tex]

has for a basis the products of open sets

[tex]\Pi_{i\in I}O_i[/tex]

where all but finitely many of those O_i are not the whole X_i.

---

say I is countable for simplicity and consider x=(x1,...) and y=(y1,...) two distinct points in the product space. I don't see how we can find two ngbh of x and y that do not intersect!
 

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  • #2
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Hausdorff, given either the product topology or box topology. This has been proven here before. The converse is true as well (though mathwonk remarked that the product must not be empty).
 
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quasar987
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I found the thread thx.
 
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quasar987
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But it seems to deal only with the converse.

Given any 2 open set of the basis containing x and y resp., since infinitely many O_i are X_i in both ngbh, they will always have a huge intersection. Is this not the case?
 
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morphism
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Ask yourself this question: Is there anything in [itex]\emptyset \times X \times X \times \cdots[/itex]?
 

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