Hausdorffness of the product topology

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SUMMARY

The discussion confirms that the product of an infinite number of Hausdorff spaces is not Hausdorff. Specifically, when considering the product topology on the space \(\Pi_{i\in I}X_i\), it is established that for any two distinct points \(x=(x_1,...)\) and \(y=(y_1,...)\), any neighborhoods of these points will intersect due to infinitely many open sets being equal to the entire space \(X_i\). This conclusion is supported by previous discussions and highlights the necessity of the product being non-empty for the converse to hold true.

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quasar987
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[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

\Pi_{i\in I}X_i

has for a basis the products of open sets

\Pi_{i\in I}O_i

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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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).
 
I found the thread thx.
 
Last edited:
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?
 
Ask yourself this question: Is there anything in \emptyset \times X \times X \times \cdots?
 

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