Topology: is this an open cover of an unbounded subspace of a metric space?

In summary, the collection {A^B(b,k)} over all natural numbers k is an open cover of the unbounded subspace A in the metric space (X,d). This is because each B(b,k) is a basis element for the metric topology on X, and therefore open in X. Since there are infinitely many balls, they will cover A. The only potential issue is if A is unbounded, but even in that case, for any specific point a in A, there is a real number k>R such that a is contained in B(b,k). Thus, the collection covers all elements of A.
  • #1
fraggle
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Homework Statement



Suppose A is an unbounded subspace of a metric space (X,d) (where d is the metric on X).
Fix a point b in A let B(b,k)={a in X s.t d(b,a)<k where k>0 is a natural number}.

Let A^B(b,k) denote the intersection of the subspace A with the set B(b,k).

Then the collection {A^B(b,k)} over all k in the natural numbers (i.e keeps going on and on) is an open cover of A in X.


Homework Equations


I believe that this is an open cover. If it is then the proof I'm doing will work.
If somebody can prove me wrong could they please suggest an open cover?
Thank you very much


The Attempt at a Solution


since each B(b,k) is a basis element for the metric topology on X B(b,k) is open in X and thus A^B(b,k) is open in A. Since there are infinitely many balls they will cover A. Therefore {A^B(b,k)} is an open cover of A as a subspace of X.

The only possible problem I see is that A is unbounded. Perhaps this collection does not cover all elements of A?

Thank you for the help
 
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  • #2
Even if A is unbounded, for any specific point in a, you have that d(a,b)=R for some real number. You can find k>R, and a is contained in B(b,k)

You might be interested in noting that b doesn't even need to be in A, just in X
 
  • #3
thanks, that makes sense
 

1. What is an open cover in topology?

An open cover in topology is a collection of open sets that completely cover a given space. In other words, every point in the space is contained in at least one of the open sets in the cover.

2. How is an open cover related to an unbounded subspace?

In order for an open cover to be considered an open cover of an unbounded subspace, the open sets in the cover must also be unbounded. This means that the open sets extend infinitely in all directions and do not have any boundaries or edges.

3. Why is it important for an open cover to be of an unbounded subspace?

In topology, the concept of an open cover is used to define a topological space. If the open cover is not of an unbounded subspace, it may not accurately represent the topological properties of the space and could lead to incorrect conclusions or results.

4. How is an open cover of an unbounded subspace different from an open cover of a bounded subspace?

An open cover of an unbounded subspace has the property that the open sets are unbounded, while an open cover of a bounded subspace has the property that the open sets are bounded. This means that the open cover of an unbounded subspace will contain infinitely many open sets, while the open cover of a bounded subspace will only contain a finite number of open sets.

5. Can an open cover of an unbounded subspace also be an open cover of a bounded subspace?

Yes, an open cover of an unbounded subspace can also be an open cover of a bounded subspace. However, the opposite is not true. An open cover of a bounded subspace cannot also be an open cover of an unbounded subspace because the open sets in the cover would have to be both bounded and unbounded, which is not possible.

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