I am attempting to figure out the following question.
Q. Let X be the normed space of bounded real sequences and norm $\| x \|_{\infty} = Sup_{1 \leq n} |x_n|$, $x=(x_n)\in C_0$. Which of the following subsets are closed in X?
A={increasing sequences}
B={sequences converging to zero}
C={convergent sequences}
D={sequences having 0 as a limit point}
E={periodic sequences}
Attempt at solution. At first glance, my thoughts are that A,B,C, and D are all closed, and that E would not be closed. My thought process is that for a space to be closed, it must contain its limit. If we consider the sequence of subsequences, say in A and B, the limit of the sequence of subsequences will be an increasing sequence and a sequence converging to $0$ respectively. However, this reasoning feels a little off to me and I was wondering if anyone could verify if (a) my initial intuition was indeed correct and (b) whether my reasoning is sensible, or suggest another option (I considered using Cauchy sequences and proving whether or not they have a convergent subsequence, but I wasn't sure how to use it in most cases).
Thank you!
What subsequences? I think you're a bit confused because we're talking about sequences of sequences. It may be helpful to think of these as two-dimensional arrays $x_{ij}$, where the $i$'th sequence $X_i$ is $x_{i,1}, x_{i,2}, x_{i,3},\ldots$, (the $i$'th row of the array) and the limit is $Y = (y_1, y_2, \ldots)$ where $y_j = \lim_{i \to \infty} x_{ij}$, the limit being uniform. Thus for (A), you could have a situation where each $X_i$ is an increasing sequence (so $x_{i,1} < x_{i,2} < \ldots$) but $Y$ is not: you might have, say, $y_1 = y_2$.
(E) is the tricky one. Think of a situation where each $X_i$ is periodic but with different periods (say period $2^i$): to get $X_{i+1}$ you change every $2^{i}$'th entry of $X_i$ slightly to spoil the $2^i$-periodicity (i.e. so $x_{i+1,j} \ne x_{i+1,j+2^i}$), but keep the $2^{i+1}$-periodicity ($x_{i+1,j} = x_{i+1,j+2^{i+1}}$).
Another way to think of this would be to see whether the complement is open. Thus for (C), if $x$ is not convergent, is there some $\epsilon > 0$ such that every sequence $y$ with $\|x - y \| < \epsilon$ is non-convergent?