# Tagged Questions

3answers
58 views

### Sequential compactness in $\mathbb{R}$

Well known result: Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuous and let $K$ be a compact set. Then, $f(K)$ is compact. I can prove this using the definition of compactness (finding a ...
1answer
17 views

### Subspace of certain series in a Hilbert space is compact

Let $E$ be a Hilbert space and let $\{x_{n}\}$ be an orthonormal basis.  Let $\{c_{n}\}$ be a sequence of positive numbers such that $\sum c_{n}^{2}$ converges.  Let $C$ be the subset of $E$ ...
1answer
16 views

### Help undestanding compactness with convergent subsequences

One way to define compactness in metric spaces is to note that in compact metric space each sequence has a convergent subsequence. Understanding compactness is difficult for me from this ...
1answer
50 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 ...
1answer
90 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 ...
2answers
387 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 ...
1answer
103 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 ...
1answer
169 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. ...
1answer
217 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 $X$ ...
2answers
273 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 $\sup(\sin(n))$?
1answer
778 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 $l$.
1answer
141 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 $\{f_n(z)\}$ ...
1answer
219 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 ...
1answer
124 views