# Compactness in a metric space

• November 16th 2007, 09:19 PM
Compactness in a metric space
Problem:

Let X = the set of real number sequence $\{ X_{n} \} ^{ \infty }_{n=1}$ such that $\sum ^{ \infty } _{n=1} |X_{n}| < \infty$

Define $d( \{ X_{n} \} , \{ Y_{n} \} ) = \sum ^{ \infty } _{n=1} |X_{n}-Y_{n}|.$

Let $A= \{ \{ X_{n} \} \in X : \sum ^{ \infty } _{n=1} |X_{n}| \leq 1 \}$

Let $N \in$ N, let $T= \{ (x^{k} \in X : k \geq N \Rightarrow x^{k} = 0 \}$

Define $\hat {T} = T \cap A$, is $\hat {T}$ compact?

My proof:

First, we claim that there exist a subsequence of every sequence in $\hat {T}$ that converges in X.

Suppose that $\{ X_{n} \}$ is a sequence in $\hat {T}$, note that $\sum ^{N} _{k=1} |X_{n}^{k}| \leq 1$

Now, $\{ X_{n} \} = \{ X_{n} ^{1}, X_{n}^{2} , X_{n}^{2} , X_{n}^{3} , ... , X_{n}^{k-1}, 0 , 0 , 0 , . . . , 0 \}$

Note that $\sum ^{N} _{n=1} |X_{n}^{1}| \leq 1$, so $\{ X_{n}^{1} \}$ is bounded. By a theorem, we know that there exist a subsequence $\{ X_{n_{i}}^{1} \}$ of $\{ X_{n}^{1} \}$ such that $X_{n_{i}}^{1}$ converges to a point in R, denote by $X_{0}^{1}$.

Now, consider the component $\{ X_{n_{i}}^{2} \}$ in the subsequence $\{ X_{n_{i}} \}$, which is also bounded. Thus there exist a subsequence $\{ X_{n_{i_{p}}}^{2} \}$ of $\{ X_{n_{i}}^{2} \}$ such that $X_{n_{i_{p}}}^{2}$ converges to a point in R, this time denote by $X_{0}^{2}$.

Note that the component $\{ X^{1}_{n_{i_{p}}} \}$ is a subsequence of $\{ X^{1}_{n_{p}} \}$, thus it also converges to the point $\{ X_{0}^{1} \}$.

Continue in the manner, we find a subsequence of $\{ X_{n} \}$, denote by $\{ X_{n_{q}} \}$ for simplification purpose, such that $X^{k}_{n_{q}} \longrightarrow X^{k}_{0} \\\\\ \forall k \in$N.

Now, consider $\lim _{q \rightarrow \infty } d( X_{n_{q}}, X_{0} ) = \lim _{q \rightarrow \infty } \sum ^{N} _{k=1} |X_{n_{q}}^{k} - X_{0}^{k} | = 0$. Thus proves that $\{ X_{n_{q}} \} \longrightarrow \{ X_{0} \} \in X$.

Now, we claim that $\{ X_{0} \}$ is a point in $\hat {T}$.

Assume to the contrary that $\{ X_{0} \}$ is not a point in $\hat {T}$. We have $\sum ^{N}_{k=1} |X_{0}^{k}| > 1$

Note that $\sum ^{N}_{k=1} |X_{n_{i}}^{k}| \leq 1$ since it is a sequence in $\hat {T}$.

But, $\sum ^{N}_{k=1} |X^{k}_{n_{i}} - X_{0}^{k} | > 0$, in which contradicts our assumption. Therefore we show that $\{ X_{0} \} \in \hat {T}$, thus proves $\hat {T}$ is compact.

Q.E.D.

Please check my proof, thank you!
• November 17th 2007, 01:14 AM
Opalg
Yes, that argument looks correct.

Another approach would be to look at the map $f: T\to \mathbb{R}^N$ defined by $f(x) = (x_1,x_2,\ldots,x_N)$, where $x = (x_1,x_2,\ldots,x_N,0,0,\ldots)$. This map is a homeomorphism from T (with the metric d) to $\mathbb{R}^N$ with the euclidean metric. The set f(T∩A) is a closed, bounded subset of $\mathbb{R}^N$, hence compact, from which it follows that T∩A is also compact.