# Question about two definitions of local compactness.

I was looking at the wikipedia definition for local compactness, and it gives various definitions, I am interested in these two:

definition A:

We say a topological space $X$ is locally compact if every point $x$ has a compact neighbourhood.

definition B:

We say a topological space $X$ is locally compact if every point $x$ has a compact local base.

The definition of a compact neighborhood of $x$ is a neighborhood of $x$ that is contained inside a compact set.

The definition of a compact local base of $x$ is a local base consisting of compact neighborhoods.

I think definition $A$ implies definition $B$:

Suppose $x\subseteq U \subseteq K$ is a compact neighbourhood of $x$. Then we can find a compact basis of $x$ by considering the neighbourhoods $U_\alpha \cap U$ with $\alpha \in A$ where the $U_\alpha$ are any basis of $x$.

Is this correct? The wikipedia page says they aren't equivalent so I am confused.

• What you did is fine. – José Carlos Santos Sep 14 '18 at 18:42
• It seems they're equivalent though – Rodrigo Dias Sep 14 '18 at 18:59
• First, you should edit your question; there are plenty of cases where you write "compact" instead of "locally compact". Second, the intersection of a compact set and an open set is generally not compact. – Michael Greinecker Sep 14 '18 at 19:12
• Oh thanks for noticing the mistake. Also, what is the point of your second sentence? – HereToRelax Sep 14 '18 at 19:36
• It all depends on your definition of neighbourhood: there are authors who define $U$ is a neighbourhood of $x$ to mean: there is an open set $O$ with $x\in O\subseteq U$. If you then take `compact neighbourhood' literally - the neighbourhood itself is compact - then A and B are not automatically equivalent. – hartkp Sep 14 '18 at 20:00

Here is an example of a compact $T_1$-space that is locally compact A but not locally compact B, where you want the neighbourhoods themselves to be compact, rather than just contained in some compact set. Take $\mathbb{Q}\cup\{\pi\}$ topologized as follows: rational numbers get their normal neighbourhoods; the neighbourhoods of $\pi$ are the co-finite sets. This space is locally compact A because the whole space is compact. It is not locally compact B, because no rational point has a neighbourhood that is compact in its own right that is disjoint from $\{\pi\}$.