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I want to find out if the space defined here is paracompact. I think it is not but I'm not able to prove it. I tried applying the fact that a Lindelöf T3 Space is Paracompact but this space isn't Lindelöf since the open cover containing only one $U_n$ would need a subcover to have uncountably many open nbds of the form $V_x \setminus A$ to cover the $x$ axis elements. I also tried proving it fully normal. I also tried applying this: "A regular space is paracompact if every open cover admits a locally finite refinement. (Here, the refinement is not required to be open.)"

I think for an open cover with a single $U_n$, any open nbd of $x(x\leq n)$ of the form $V_x\setminus A$ or $U_m$ has to intersect infinitely many members of the refinement otherwise it would contradict the fact that each $V_y\setminus A$ has only finitely many holes. Kindly provide hints.

ZSMJ
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    Paracompact Hausdorff spaces are normal, hence completely regular. – Ulli Sep 26 '22 at 09:02
  • @which gives us a contradiction, thanks!. Do you think proving this straight from the definition of Para compactness is also trivial? – ZSMJ Sep 26 '22 at 11:43
  • You could try to adapt the proof that paracompactness implies normality to this space. But then you still have to show that the space is not normal. – Ulli Sep 26 '22 at 12:43

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