0
votes
1answer
42 views

Compact subsets of $c_0$

Let $c_0$ be the Banach space of all sequences converging to 0, equipped with the supremum norm. How do the compact subsets of $c_0$ look like? I could imagine that $K \subset c_0$ is compact if ...
2
votes
1answer
79 views

uniformly convergent sequence of functions on a compact space

There's an exercise on Kaplansky's textbook that says: Let $\{f_i\}$ and $f$ be continuous real functions on a compact metric space $M$. Prove that $f_i$ converges uniformly to $f$ if and only if the ...
1
vote
2answers
296 views

Sequence has a convergent subsequence in R^n

Suppose A is a closed and bounded subset of R^n. Let {ak} be a sequence in A. Thus, the elements of {ak} are: (a11,a12,...,a1n), (a21,a22,...,a2n), ... ... (ak1,ak2,...,akn), ... We are not sure if ...
0
votes
1answer
102 views

Working in $\mathbb{R}^n$. Prove that there exists $x$ in $A$ and $y$ in $B$ such that $\mathrm{dist}(A,B)=\|x-y\|$.

Let $A$ be a compact subset of $\mathbb{R}^n$ and let $B$ be a closed subset of $\mathbb{R}^n$. Also suppose $A$ and $B$ are disjoint. Prove that there exists $x$ in $A$ and $y$ in $B$ such that ...
0
votes
1answer
157 views

Real analysis, showing that a set is not compact.

I am limited to theorems from Rudin, so think basic real analysis. Given: $l^1$ - set of sequences such that the infinite series consisting of terms $|a_n|$ converges. i.e absolute convergence. ...
3
votes
1answer
194 views

Let $X$ be a compact metric space. If $f:X\rightarrow \mathbb{R}$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum.

Let $X$ be a compact metric space. If $f:X\rightarrow \mathbb{R}$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum. I want to prove this. This is my proof: Since $X$ ...
1
vote
2answers
231 views

What is $\sup(\sin(n))$? [duplicate]

My classmate asked a question during lecture about our discussion of bounded sequences, particularly the sequence $\sin(n)$. His question was, What is $\sup(\sin(n))$?
1
vote
1answer
637 views

If every convergent subsequece converges to the same limit then the sequence converges

I came across this question: In a compact metric space $(X,d)$ if every convergent subsequence of a sequence converges to the same limit, say $l$, then the original sequence also converges to $l$.
2
votes
1answer
126 views

I don't understand this remark regarding Weierstrass' Theorem (Ahlfors' Complex Analysis)

In Ahlfors' text of Complex Analysis, chapter 5 theorem 1, he proves the following: Theorem I. Suppose that $f_n(z)$ is analytic in the region $\Omega_n$, and that the sequence $\{f_n(z)\}$ ...
2
votes
1answer
199 views

The set of all subsequential limits of a bounded sequence is a non-empty compact set

Let $(x_n)$ be a bounded sequence and let $Y$ be the set of all subsequential limits of $(x_n)$. Prove that $Y$ is a non-empty compact set. I think it's possible to solve this problem by proving that ...
0
votes
1answer
113 views

Countable product of finite sets with a new metric, compact?

Suppose we have a finite set $E$. Is it true that $E^n$ is compact? The metric on $E^n$ is : $$d(\omega,\omega\prime)=\begin{cases}2^{-\inf \{ n \in \mathbf N:\omega _n \ne \omega'_n\} }&{\omega ...
4
votes
2answers
793 views

Closed Bounded but not compact Subset of a Normed Vector Space

Consider $\ell^\infty $ the vector space of real bounded sequences endowed with the sup norm, that is $||x|| = \sup_n |x_n|$ where $x = (x_n)_{n \in \Bbb N}$. Prove that $B'(0,1) = \{x \in l^\infty ...
5
votes
2answers
638 views

Unit ball in $C[0,1]$ not sequentially compact

This question is taken from Saxe K -Beginning Functional Analysis. Show that the closed unit ball in $C[0,1]$ is not compact by proving that it is not sequentially compact. (It's assumed that we ...
5
votes
2answers
165 views

Closed set in $\ell^1$

Show that the set $$ B = \left\lbrace(x_n) \in \ell^1 : \sum_{n\geq 1} n|x_n|\leq 1\right\rbrace$$ is compact in $\ell^1$. Hint: You can use without proof the diagonalization process to ...
1
vote
0answers
201 views

Example of compactness of the set of elements of a convergent sequence.

An example in a book I am reading on general topology demonstrates the concept of compactness by showing that the set $E = \{s_n : n = 0,1,2,3 \ldots \}$ is compact in some topological space S. The ...
0
votes
1answer
143 views

Yes or No, Real Analysis, continuity, compactness

Am I correct over statements below? The limsup and liminf of the sequence $n^2$ (meaning $1,4,9,16,\dots$) are equal. T Every bounded sequence has at most one ...
2
votes
2answers
79 views

The compactness of $\{x_n\}_{n\in\mathbb{N}}=\{\sin nt\}_{n\in\mathbb{N}}\in L_2[-\pi,\pi]$

Is the set $\{x_n : n\in\mathbb{N}\}=\{\sin (nt): n\in\mathbb{N}\}\subset L_2[-\pi,\pi]$ closed, or further compact? I don't know how to prove it.
4
votes
1answer
160 views

Prove that the following subset is compact

Let $A \subset \ell^p$, where $1 \le p \lt \infty$. Suppose the following conditions are true: 1) $A$ is closed and bounded 2) $\forall \epsilon \gt 0, \: \exists \: N \in \mathbb{N}$ ...
3
votes
1answer
183 views

Compact subsets in $l_\infty$ (converse of my last question)

(Converse of my last question) If $A \subseteq \ell_\infty$, and $A=\{l\in \ell_\infty: |l_n| \le b_n \}$, where $b_n$ is a sequence of real, non-negative numbers, then if $\lim (b_n) = 0$ it must ...
2
votes
2answers
74 views

characterisation of compactness in the space of all convergent sequences

I go through a proof of the following. Let $(\ell_1,d)$ be the metric space of all sequences $x = (\xi_i)_{i \in \mathbb{N}}$ with $\sum_{i=1}^{\infty} |\xi_i| < \infty$ and the metric $$ d(x,y) ...