Let $(c_0,\|\space.\|_{\infty})$ be the space of sequences $(x_k)_{k=0}^{\infty}$ such that $x_k\in{}\mathbb{K}$ and $x_k\rightarrow{}0$. I want to show that this space is complete.  I understand this requires me to take a Cauchy sequence in $c_0$ and showing it not only converges, but converges to a limit point (a sequence) in $c_0$.  I am comfortable with the definitions, but I am not sure how to approach the proof. This is how I tried to break down the problem:  Clearly a Cauchy sequence is bounded and I believe it does converge but can't show why (I can easily show that a convergence implies Cauchy).  After this, knowing it converges, I must show it converges to a sequence that itself converges to 0.  I did find a similar question someone else has posted but didn't find the solution very helpful.
120404
Let $ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$. Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$  I am capable of showing the space where the limit of $x_n$ exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0.   Let $(x^{(n)})_{n \in \mathbb N}$ be a Cauchy sequence in $c_0$ such that $x^{n} = (x^n_1, x^n_2,...)$. Fix $k \in \mathbb N$ consider the sequence $(x^n_k)_{n \in \mathbb N}$ in $\mathbb F$. For any $n,m \in \mathbb N$  $\lvert x^n_k - x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k - x^m_k \rvert = \lVert x^n - x^m \rVert_\infty \lt \epsilon $   (1)  Thus $x^n_k$ is Cauchy in $\mathbb F$ and so has limit $y_k$ such that $y = (y_1,y_2,...)$ and y is the limit of $x^n$  To show that such a y exists we look at the value of $\lvert y_n - y_m \rvert \le \lvert y_n - x^N_n \rvert + \lvert x^N_n - x^N_m \rvert + \lvert x^N_m - y_m \rvert \lt \epsilon$ for all $n,m \ge N$     (2)  The middle expression on RHS of (2)  is $\lt \epsilon/3$ by (1)  The other two are also $\lt \epsilon/3$ follow from $x^N_k$ being Cauchy and converging to $y_k$  This shows that $\lim_{n \to \infty} y_n$ exists but we still have not shown that $y \in c_0$.  I know that to show y tends to 0 i should show that $\lvert y_k \rvert \lt \epsilon$ for $k \ge N$  This is where I am stuck. Perhaps $\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$ and then we can take the limit function outside the absolute value sign by continuity? Then we might say due to it being a Cauchy sequence $x^n_k \lt \epsilon$. I know this last bit isn't at all convincing so I could do with some help.
68767
Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \  a_n, b_n \neq 0$$  Show that $(\forall  k \in \mathbb N^*,  \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$  The vector-space should be a finite-dimensional $\mathbb R$-vector-space.   I do not see how to deal with the hypothesis. If you have ideas/hints...  It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases.
53269
Let $(x_n)$ be a sequence of real numbers.  Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .              Thanks for any help.
75160
Let $(y_n)\subseteq\mathbb{C}$ a sequence such that $\sum_{n=1}^{\infty}x_n\overline{y_n}$ converges for every $(x_n)\in c_0$. ($c_0$ is the space of complex sequences converging to zero with the $\infty$-norm.) Show that  $\sum_{n=1}^{\infty}|y_i|&lt;\infty$. Hint: uniform boundedness theorem.   Since $c_0$ is a Banach space, in order to use the uniform boundedness theorem we need linear bounded operators $T_n:c_0\to \mathbb{C}$. My first guess was:  $$T_n(x)=\sum_{i=1}^nx_i\overline{y_i}$$  Clearly every $T_n$ is linear but to prove $T_n$ is continuous seems to be the same as to prove what we actually desire.   Would anyone give me a hint about how to define the family of bounded operators, in order to use the theorem?  Thank you.
8578
Let $a_n \subset \mathbb C$ is a sequence Show that if $\sum_{n=1}^{\infty} a_nx_n \lt \infty$ $\forall x_n \in \ell^1$   then $a_n \in \ell^{\infty}$  What did I do : $x_n \in \ell^1 \Rightarrow \sum_{n=1}^{\infty} |x_n| \lt \infty $ Since $\mathbb C$ is a Banach space $\sum_{n=1}^{\infty} x_n \lt \infty $ and from Leibnitz rule $x_n \to 0$ $\sum_{n=1}^{\infty} a_nx_n \lt \infty \Rightarrow$ $a_nx_n \to 0$ i.e. $\forall \varepsilon \gt 0$$ |a_nx_n| \lt \varepsilon $ $|a_n||x_n|=|a_nx_n| \lt \varepsilon \Rightarrow$ $|a_n| \lt \varepsilon /|x_n| $ and $|x_n|$ is a finite number thus terms of $a_n$ are bounded. So $a_n \in \ell^{\infty}$ I’m not sure about my writtens. It seems wrong to me. Could someone fix me please? Thanks a lot
110984
Suppose $E$ is a normed space of all continuous functions on $[0,1]$ with norm $\int_{0}^{1}|f(t)|dt$. Prove that $E$ is not complete   I know that we must do is to find a Cauchy sequence of continous functions that doesn't converge in $E$, but I can't find that sequence. Can any one help me? Thanks.
115625
$X$ is a normed vector space. Assume $X$ is reflexive, then $X$ must be a Banach space.  I guess we only need to show any Cauchy sequence is convergent in $X$.
31840
(Taken from an old exam - no homework:)      Prove that every convergent sequence of real numbers is bounded.   It would be cruel if our professor would take a task like that in the exam because I rather prefer calculating than doing theory (in an exam where you haven't got much time to describe things..). But I'll be honest and say I dislike theory in general! :-)  Anyway, for the task, I really got no idea how to proof that but I'll give it a try...    We know that if a sequence is convergent it has exactly ONE limit where the sequences (sry my English here) accumulate. So the sequence will have an upper bound or lower bound, based on the limit, thus it will be surely bounded if it's convergent sequence.    Even if my proof is wrong (most likely..) I'd like to see a proof like mine, where you avoid using (complicated) maths symbols. Just using words any human can understand would be awesome! :-)  Of course a more formal proof is welcome too but not necessarily preferred ^.^
71569
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes:  Suppose you have a sequence $a_n$ of real numbers such that $$\lim_{n\to\infty} a_{n} = 0$$ and suppose the sequence of partial sums $s_n$ is bounded. Prove that $s_n$ converges or give a counterexample.  I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated.
22048
Put   $$ l^{\infty} = \{ (x_n) \subseteq \mathbb{C} : \forall j \; \;\ \;|x_j| \leq C(x)\} $$  I want to show that $c_0$, the space of all sequences of scalars that converges to $0$ is closed subspace of $l^{\infty}$.  MY atttemt: Take arbitrary sequence $(x_n) \in c_0$. We know $x_n \to 0$. but $0 \in c_0$. Therefore, every sequence in $c_0$ converges to an element in $c_0$. Hence, $c_0$ must be closed  Is this correct? thanks
57190
Let $\{x_n\}$ be a sequence such that there exists a $0 &lt; C &lt; 1$ such that $$|x_{n+1} - x_n| \le C |x_n - x_{n-1}|.$$  Prove that $\{x_n\}$ is Cauchy. Hint: You can freely use the formula (for $C \not = 0$) $$1+C+C^2+...+C^n = \frac {1-C^{n+1}}{1-C}.$$  My attempt : $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}&lt;C$. Then, $\frac {|x_{n+1} -x _n|}{|x_n - x_{n-1}|}&lt;1$. So, I thought that it goes to zero as $n \to \infty$ because the difference between $x_n$ and $x_{n+1}$ is getting smaller as $n \to \infty.$ But, I don't know how to use the hint in the question, and how to write the answer properly.  Could you give some hint?
88024
I am trying to understand better Cauchy sequence in the book that I am reading he is trying to provide counter argument that in general metric spaces we can find sequence that are Cauchy but aren't convergent for example $\frac{1}{n}$ is Cauchy but doesn't converge in the space T = (0,1].  I am trying to prove that the sequence is Cauchy to begin with if we consider $n \geq m &gt; 0$ Consider $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = 1/n &lt; \epsilon$. So if we choose N = $\frac{1}{\epsilon}$ that should work since for $n, m &gt; N = \frac{1}{\epsilon}$   we have the following:  $|a_n - a_m| = \frac{m - n}{mn} &lt; \frac{m}{mn} = \frac{1}{n} &lt; \epsilon$.  what do you guys think any the proof is good or is there anything to make it better ? It is always good to learn from the extra bit of details as it makes my brain understand the subject more !
36376
Let $C=C[0,1]$ be the space of all continuous functions on $[0,1]$. Define $\|f \|=\max \ |f(x)|$. I want to show that $C$ is a Banach space.     Below is my attempt and I was wondering if it's ok.    I know I have to show that $C$ is a complete normed space. Clearly, $\|f\| \geqslant 0$ and $\|f\|=0 \Leftrightarrow f=0$. $\|cf \|=\max~|cf(x)|=|c|\max |f(x)|=|c| \cdot \|f\|$. $\|f+g\|=\max~|f(x)+g(x)|\leq \max~|f(x)|+\max~|g(x)|=\|f\|+ \|g\|$. So $C$ is a normed space.   Next, I show that every Cauchy sequence in $C$ is convergent. Let $\{f_n\}$ be a Cauchy sequence in $C$. Let $\varepsilon \gt 0.$ Then $\exists$ an $N_1$ such that $$ \max~|f_n(x)-f_m(x)| \lt \frac{\varepsilon}{2}$$ for $n, m \gt N_1$ and $x\in[0,1]$.  But there is a subsequence $f_{k_n} $, which converges to $f$. So $\exists$ an $N_2$ such that $$ \max~\left|f_{k_n} - f\right|\lt \frac{\varepsilon}{2}$$ for each $n\gt N_2$.  Now Let $N = \max\{N_1, N_2\}$, if $n \gt N$ then $k_n \geqslant n\gt N$. So we have $$ \max~\left|f_n(x) - f(x)\right| \leqslant \max~\left|f_n - f_{k_n}\right| + \max~\left| f_{k_n} - f\right| \lt\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$  Thus, $\|f_n-f\| \to 0$ as $n\to \infty$. $\quad \square$  Thanks.
9450
Let $(X,d)$ be a metric space, and say $(x_n)$ is a Cauchy sequence such that it has a convergent subsequence $(x_{n_k})$ that converges to $x$. We show $x_n \to x$. Let $\epsilon &gt; 0$. Take $N &gt;0$ such that for all $n,m &gt; N$, we have   $$d(x_n,x_m) &lt; \frac{\epsilon}{2}.$$  By hypothesis, we can take also $K &gt;0$ such that for all $n_k &gt; K$, we have  $$ d(x_{n_k},x) &lt; \frac{\epsilon}{2}.$$  Put $M = \max \{N,K\}$. Therefore, for all $n,m,n_k &gt; M$, we have  $$ d(x_n,x) \leq d(x_n, x_{n_k}) + d( x_{n_k},x) &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$  Hence, $x_n \to x$ as desired.  Is this a correct approach? Thank you very much in advance.
41734
Consider the Banach space $X$ of null sequence whose elements are complex sequence which converges to $0$. In addition the norm is defined as $$\|(a_1, \dots, a_n)\| := \sup_n |a_n|.$$ Show this space is NOT reflexive. Recall that the dual of $X$ is $l_1$. Also use the following fact. If $X$ is reflexive space and $(x_n)$ is a sequence in $X$ and for all $\psi \in X^*$, the sequence $\psi(x_n)$ has a limit in $\mathbb C$, then $x_n$ converges weakly to some $x \in X$.   Suppose that $X$ is reflexive. Consider the sequence $$u_n = (1, \dots, 1, 0, \dots) \in X,$$ where first $n$ entries are $1$. Then for any $y \in l_1$, we have $$y(u_n) := \sum_n y_n\times u_n = y_1+\dots+y_n \to \sum_n y_n &lt; \infty,$$ since $y$ is absolutely convergent. Now it is only left to show that $u_n$ does not converges weakly to some $u \in X$ which I guess is $(1, 1, \dots)$. That is, for all $y \in l_1$, $$y(u_n) \nrightarrow y(u).$$ However, this seems correct to me. Where did I make mistake, please? Thank you!
105067
In the $ US $ space , each convergent sequence has unique limit.      In the $ KC $ space , every compact subset is closed.      It easy to show that $ KC $ spaces imply $ US $ spaces.   The example below show that $ US $ spaces do not imply $ KC $ spaces.      Let $ X = [ 0, 1 ] $ with the the usual topology and attach  a new point z whose neighborhood are open dense subsets of $ [ 0,1 ]$. Observe $ [ 0,1 ]$ is a compact non-closed  of $X$ and then $X$ is not a $KC$ space. However no sequence in $ [ 0,1 ]$ converge to $z$ and in particular  all convergent in $X$ have unique limits.   can you give me more details?   Why $X$ is not closed? Why no sequence in $ [ 0,1 ]$ converges to $z$ and in particular all convergent sequences in $X$ have unique limits ?
62673
I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.  What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?  By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not.