Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
Definition: The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.

Using these definitions, prove that int(F) is an open set.

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Attempt:
Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

Any help is greatly appreciated!

Originally Posted by kingwinner

Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
Definition: The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.

Using these definitions, prove that int(F) is an open set.

============================

Attempt:
Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

Any help is greatly appreciated!

As a very brief hint, try taking r' = r/2 and using the triangle inequality.

Originally Posted by kingwinner

Let (X,d) be a metric space and let F be a subset of X. B(r,x) is the open ball of radius r about x.
Definition: The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.

Using these definitions, prove that int(F) is an open set.

============================

Attempt:
Let x E int(F), then there exists r>0 such that B(r,x) is contained in F.
We need to prove that there exists some r' >0 s.t. B(r',x) is contained in int(F). How can we prove this?

Any help is greatly appreciated!

That is a strange definition for the interior. The way that I have most often seen it defined as is the union of all open subsets of the set.

Originally Posted by Opalg

As a very brief hint, try taking r' = r/2 and using the triangle inequality.

So you mean B(r/2,x) is contained in int(F)? How can we prove this? What is the intermediate point in the triangle inequality?

thanks.

Originally Posted by kingwinner

So you mean B(r/2,x) is contained in int(F)? How can we prove this? What is the intermediate point in the triangle inequality?

If $\displaystyle y\in B(r/2,x)$ then $\displaystyle B(r/2,y)\subseteq B(r,x)\subseteq F$, because $\displaystyle d(z,y)<r/2$ and $\displaystyle d(y,x)<r/2$ together imply that $\displaystyle d(z,x)<r$, by the triangle inequality. Therefore $\displaystyle B(r/2,x)\subseteq \text{int}(F)$.