Talk:Compact in Subspace is Compact in Topological Space

Nice work. I removed the condition that $K$ is compact, because it is not needed (and not used in the proof). I hope you're fine with that. --barto (talk) 14:16, 31 August 2017 (EDT)

You are right. Thank you.--Bltzmnn.k (talk) 14:32, 31 August 2017 (EDT)