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Find an uncountable number of subsets of $\ell_{n}^{p}(\mathbb{R})$ and $\ell_{n}^{p}(\mathbb{C})$ that are neither open nor closed.

Attempt: For $\ell_{n}^{p}(\mathbb{R})$ , take the collection of sets $A_n=(n,n+1$] with $\mathbb{R}$ as the index set.

For $\ell_{n}^{p}(\mathbb{C})$ , since $\mathbb{R\subset\mathbb{C}}$, wouldn't the same subsets I use for $\ell_{n}^{p}(\mathbb{R})$ work?

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What's $l^p_n$? –  Matt N. Jan 31 '12 at 20:39
$\ell_{n}^{p}(\mathbb{R})$ is the space ($\mathbb{R^n},d_p$) where $d_p(x,y)= \max_{1\leq k\leq n}|x_j-y_j|,1\leq p \leq\infty$. –  Emir Jan 31 '12 at 22:54
Usually, $\ell^p$ is used to denote that the metric is $d(x,y)=\bigl(\sum |x_i-y_i|^p\bigr)^{1/p}$. What you have above is denoted by $\ell^\infty$. All these metrics induce equivalent norms, however. –  David Mitra Jan 31 '12 at 23:06

1 Answer 1

up vote 2 down vote accepted

Elements in $\ell_n^p$ are sequences with $n$-coordinates. I do not think your $A_n$ will work.

Hint: Think of closed balls with their centers deleted: $\overline B_\epsilon(x)/\{x\}$, where $x\in\ell_n^p$. None of these are open since no open ball containing a boundary point of $\overline B_\epsilon(x)/\{x\}$ is contained in $\overline B_\epsilon(x)/\{x\}$. And none are closed since $x$ is a limit point of $\overline B_\epsilon(x)/\{x\}$ not contained in $\overline B_\epsilon(x)/\{x\}$.

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I'm a little confused about the reasoning why $\overline B_\epsilon(x)/\{x\}$ is not open -- isn't a closed ball by definition a closed set? –  Emir Jan 31 '12 at 22:49
@Emir. I was showing directly that these sets were not open. By "on the boundary of $\overline B_\epsilon(x)/\{x\}$", I meant any point $y$ with $d(x,y)=\epsilon$. No open ball containing $y$ is contained in $\overline B_\epsilon(x)/\{x\}$. So this set is not open. Closed balls are closed sets in these metric spaces, but I think a bit more argument is needed... –  David Mitra Jan 31 '12 at 23:02
Ah, now I see why my original argument was insufficient. So if these balls are neither open nor closed in $\ell_{n}^{p}(\mathbb{R})$ they are also neither open nor closed in $\ell_{n}^{p}(\mathbb{C})$ since $\mathbb{R}\subset\mathbb{C}$, right? –  Emir Feb 1 '12 at 2:35

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