Metric Space and Lindelof Theorem

In summary, the conversation discusses proving that a metric subspace (X,d) must obey Lindelof if the metric space (K,d) it is a subset of also obeys Lindelof. The conversation suggests using a countable open cover of (K,d) to also cover (X,d), and potentially showing that the elements of this cover are open in the subspace. However, it is noted that it may be difficult to prove this directly and the suggestion is made to use the fact that Lindelof is equivalent to being second countable in a metric space.
  • #1

Homework Statement

Assume some metric space (K,d) obeys Lindelof, take (X,d) a metric subspace of (K,d) and show it too must obey Lindelof.

The Attempt at a Solution

I'm assuming since I know that (K,d) obeys Lindelof then there is some open cover that has a countable subcover say {Ji | i is a positive rational number}. Since this covers all of the metric space, then it surely covers the metric subspace. So I'm thinking all I need to really prove is that the elements of this set of open countable covers is open under metric subspace (X,d)? Or do I need to show that I can select a countable number of {Ji} to cover (X,d)? If so, any suggestions on how to do so?

This is the last assignment question, and it's due this Wednesday November 2nd, any help would be appreciated!
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  • #2
It will be hard (even impossible) to prove this directly. For the simply reason that the result isn't true in an arbitrary topological space.

You need to do a little shortcut. Do you know that in a metric space that Lindelof is equivalent to being second countable?? (prove it if you don't know)
Well, using this, it suffices to prove that a subspace of a second countable space is second countable. This is a lot easier.
  • #3
No we haven't learned about second countable yet, so I shouldn't be using that in my proof. Any other suggestions? Thank you!

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