Talk:Intersection of Closed Set with Compact Subspace is Compact/Proof 1

Bltzmnn.k, it seems you wanted to address the ambiguity between compactness of a space and compactness in an ambient space. But I'm afraid your edit introduces many errors:


 * The definition of the topology of $T_K$ is indirect. You just want to say that $\tau_K$ is the induced topology on $K$, but I see no need to do so.
 * The conclusion that $H\cap K$ is a union of open sets cannot be correct.

Please let us know what you think could have been improved, because it is unclear. If you want to address the fact that the above notions of compactness are equivalent (which they are), this is not the place to do so. See Equivalent Definitions of Compact Topological Subspace. --barto (talk) 13:12, 30 August 2017 (EDT)
 * I personally don't like the way the definitions of a compact subspace are currently presented. It needs to be stressed that they are equivalent. As it stands, pages such as Definition:Compact seem to imply that they are not. --barto (talk) 13:16, 30 August 2017 (EDT)


 * There is a deliberate attempt to distinguish "topological" definitions in the context of metric spaces, which are to use the language of metric spaces, from the general language of topology. This is so that the "intellectual journey" can be made from the real number Euclidean spaces, through the more general metric spaces, through to the fully general concept of "open sets", such that at each stage the continued generalisation can be shown to encompass the more specific. That is, an "open set" in a metric space is the same thing as an open set in the same space treated as a "topological" space rather than a "metric" space. Hence a compact set / subspace: its definition in the metric space is the different from what it is in its topogisation, but the sets are the same. --prime mover (talk) 15:16, 30 August 2017 (EDT)


 * Thank you for your comments. But I think you may misunderstand a little. First, I do not conclude that $H\cap K$ is a union of open sets. Instead, I show that any open cover of $H\cap K$ has a finite subcover in $T$. Second, $A\in \tau_K$ does not implies $A\in \tau$. This can be easily checked with definition of subspace.--Bltzmnn.k (talk) 22:17, 30 August 2017 (EDT)