TITLE: the product $(X_1, T_1)\times\ldots\times(X_n,T_n)$ is locally compact, then each $(X_i, T_i)$ is locally compact.
QUESTION [1 upvotes]: Prove that the product $(X_1, T_1)\times\ldots\times(X_n, T_n)$ is locally compact, then each $(X_i, T_i)$ is locally compact.
My proof: Let $p=\langle x_1,\ldots,x_n\rangle$ be in the product.
Since the product is locally compact, there exists a compact neighborhood $K_1\times\ldots\times K_n$ of $p$. Since $K_1\times\ldots\times K_n$ is compact, each factor $K_i$ is compact. 
There is an open set $U_1\times\ldots\times U_n$ in the product such that $p$ is in $U_1\times\ldots\times U_n$ which is the subset of $K_1\times\ldots\times K_n$. So, each $U_i$ is open in $X_i$. For each $x_i$ of a compact neighborhood $K_i$, there exists an open set $U_i$ for $i=1,2,\ldots,n$ such that $x_i$ is in $U_i$ which is the subset of $K_i$. Therefore each $(X_i,T_i)$ is locally compact. Is it ok?

REPLY [1 votes]: There are competing definitions of local compactness. One of them means that the space has arbitrary small compact neighborhoods. The other one means that the space is Hausdorff and each point has a compact neighborhood. The second one implies the first one.
No matter which one you are using, if the product has the property, then each factor has it. This can be seen by noting that the projection maps are open and continuous. And if $f:X\to Y$ is a continuous open surjection, $y\in Y$ is a point and $x\in X$ such that $f(x)=y$, then any compact neighborhood $K$ of $x$ maps to a neighborhood $f(K)$ of $y$ since $f$ is open and this $f(K)$ is compact since $f$ is continuous. If you want $f(K)$ to be contained in some given neighborhood $V$, just choose $K$ small enough so it is in $f^{-1}(V)$.
Also note that each factor is Hausdorff if the product is. So the statement is true for both definitions.
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