Are these two definitions for paracompactness equivalent at least for manifolds?

Question

My lecturer defined paracompactness as follows:

A topological space $X$ is called paracompact if for any open cover $\{U_\alpha\}$ of $X$ there is a locally finite subcover, i.e. for any $x\in X$ there are some $\alpha_1, ..., \alpha_k$ such that $x\in U_{\alpha_1} \cap ...\cap U_{\alpha_k}$.

Then another instructor defined paracompactness like this:

A topological space $X$ is called paracompact if for any open cover $ \{U_\alpha\}$ of $X$ there is a locally finite refinement, where locally finite and refinement are defined exactly like here.

These two definitions don't look equivalent to me simply because a subcover is a refinement, but this is not true the other way around. However, I encountered these definitions in a differential geometry course, so I am wondering whether for manifolds (which are Hausdorff and second countable) the first definition is in fact equivalent to the second one (which I am sure is the correct one since this is how I found paracompactness defined everywhere).

Answer 1

I think in the first definition you've meant

i.e. for any $x\in X$ the set $\{U_\alpha\ |\ x\in U_\alpha\}$ is finite

or maybe

i.e. for any $x\in X$ there is an open neighbourhood $U$ of $x$ such that $\{U_\alpha\ |\ U_\alpha\cap U\neq\emptyset\}$ is finite

The second variant is closer to the (as you will see soon) correct definition of paracompactness. Otherwise the definition you've proposed is satisfied by every open cover of every topological space.

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Either way, the answer is: no, these are not equivalent. For manifolds as well.

Lemma. Assume that $X$ is a topological space. Then every open cover of $X$ admits locally finite subcover if and only if $X$ is compact.

Proof. "$\Leftarrow$" Simple exercise.

"$\Rightarrow$" Let $\mathcal{U}$ be an open cover of $X$. We will show that under our assumptions it has a finite subcover.

Pick $U\in\mathcal{U}$ and $x\in U$. Define $\mathcal{V}=\{V\cup U\ |\ V\in\mathcal{U}\}$. Note that $\mathcal{V}$ is a new cover of $X$ such that $x$ belongs to every open set in the cover. By our assumption $\mathcal{V}$ admits a locally finite subcover $\mathcal{V}'$, but that is possible only when $\mathcal{V}'$ is finite (regardless of which definition we choose), because $x$ belongs to every element of $\mathcal{V}'$ as well. But then for each $V'\in\mathcal{V}'$ there is $V\in\mathcal{U}$ such that $V'=V\cup U$. These new $\{V\}$ together with $U$ form a finite subcover of $\mathcal{U}$. Thus $X$ is compact. $\Box$

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As you can see your first definition is equivalent to compactness. And therefore is wrong.