So while reading One Point Compactification this came to my mind.
Suppose there is a locally compact Hausdorff space $(X,\tau)$ then obviously it has a one point compactification,say, $Y$.Now, give the same space X a topology $\delta$ that is finer than $\tau$. Will $(X,\delta)$ have a one point compactification as well and if yes, will that be the same $Y\ ?$
To find the answer,if we could show whether or not $(X,\delta)$ is locally compact Hausdorff then our answer would be obtained.
Now,for being finer $\delta$ is obviously Hausdorff too. But what about local compactness?If it were only compact, then I know a finer topology may not retain compactness but not sure what happens when dealing with local compactness.How does Local Compactness depend on the fineness of a topology ?
A topology finer than a locally compact topology may not be locally compact.
Consider the Sorgenfrey line, that is the real numbers $\mathbb{R}$ with the topology generated by the family of half-open intervals $[a,b)$ for $a < b$. This topology is finer than the usual (metric/order) topology on $\mathbb{R}$ (if $a < b$, then $(a,b) = \bigcup_{n \in \mathbb{N}} [a+\frac{b-a}{n} , b)$). However the Sorgenfrey line is not locally compact. The answers to this question characterise the compact subsets of the Sorgenfrey line, and in particular they are countable, which implies that they have empty interior. (In terms of the basic open sets, note that for any $a<b$ we have that $[a,b) = \bigcup_{n \in \mathbb{N}} [ a , b-\frac{b-a}{n} )$, and this open cover has no finite subcover.)