Thread: "Closed" set in a metric space

1. "Closed" set in a metric space

Suppose that (x_n) is a sequence in a metric space X such that lim x_n = a exists. Prove that {x_n: n E N} U {a} is a closed subset of X.

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Let B(r,a)={x E X: d(x,a) < r} denote the open ball of radius r about a.
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.
Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F)
Theorem: F is closed in X iff the complement of F is open.

I know the definitions, but I just don't know out how to construct the proofs formally. I am not sure how to prove that some set is "closed"...I can't find any similar examples in my textbook.

May someone kindly help me out?
Any help is appreciated!

[note: also under discussion in math links forum]

2. Accidental post, see next post.

3. Originally Posted by kingwinner
Suppose that (x_n) is a sequence in a metric space X such that lim x_n = a exists. Prove that {x_n: n E N} U {a} is a closed subset of X.

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

Let B(r,a)={x E X: d(x,a) < r} denote the open ball of radius r about a.
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.
Definition: Let F be a subset of X. F is called closed iff whenever (x_n) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F)
Theorem: F is closed in X iff the complement of F is open.

I know the definitions, but I just don't know out how to construct the proofs formally. I am not sure how to prove that some set is "closed"...I can't find any similar examples in my textbook.

May someone kindly help me out?
Any help is appreciated!

[note: also under discussion in math links forum]
Let $E$ be the set in question. Consider $E'$ and let $e\in E'$. Let $\delta=\frac{d(a,e)}{2}>0$ and consider $B_{\delta}(e),B_{\delta}(a)$ these are disjoint open sets about $e,a$ respectively. Since $x_n\to a$ we have that $B_{\delta}(a)$ contains all but finitely many points of $\{x_n\}$ say $x_1,\cdots,x_m$. Let $B_{\delta_k}(e)$ be the open ball of radius $\frac{d(e,x_k)}{2}$ and consider $O=B_{\delta}(e)\cap\bigcap_{k=1}^{m}B_{\delta_m}(e )$, clearly $O$ is an open set containing $e$ disjoint form $E$. The conclusion follows.

4. In a metric space $(X,d)$ you should know that
$D(A;x)=\inf\{d(a,x):a\in A\}$ is the distance the point $x$ is from the set $A$.
Moreover, you should know that A closure $\overline{A}=\{x(A;x)=0\}" alt="\overline{A}=\{x(A;x)=0\}" />