Let $ X $ be a topological space, $ A $ be its subspace which is paracompact, and $ \{U_i\}_{i \in I} $ be an open covering of $ A $ in $ X $ (i.e., each $ U_i $ is an open subset of $ X $ and $ A \subseteq \bigcup_{i \in I} U_i $). Then, is there an open covering $ \{V_j\}_{j \in J} $ of $ A $ in $ X $ (in the same sense as above) which is locally finite in $ X $ and refines $ \{U_i\}_{i \in I} $?
Of course, since $ A $ is paracompact, there is a locally finite open refinement in $ A $, but what about in $ X $?
It seems false to me, but I could not make a counterexample.
Best Answer
As Daniel Fischer pointed out in the comments, we can make a counterexample by adding a point to $ A $ which has no non-trivial neighbourhood.
Even if $ X = [0, 1] $, there is still a counterexample. Let $ A = [0, 1) $ and $ U_x = [0, x) $ for $ 0 < x < 1 $. Then $ \{U_x\}_{0 < x < 1} $ is an open covering of $ A $ in $ X $, but has no refinement which covers $ A $ and locally finite at $ 1 $.