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I was reading through general-topology posts and I couldn't quite understand this one. I tried asking directly on the thread, but I didn't get a response.
This is in reference to Professor Israel's answer on How can I show this set of sequences is compact?. I followed all the logic up until the last sentence where he claims:

That is, $\iota(B)$ is the intersection of the closed sets $K(a,x,b,y)=\{p:ap_x+bp_y\}$ for $x,y\in\ell_1$ and $a,b\in\mathbb{R}$, so it is closed.


I'm probably missing something obvious, but why is $\iota(B)$ the intersection of those closed sets? The way I'm interpreting the claim is that $\iota(B)$ is a subset of $\mathbb{R}^{\ell_1}$ while the $K$'s are just a set of linear maps (if I understood that notation correctly) so $\iota(B)$ can't possibly be the intersection of them, much less compared with them since one is a set of elements and the other is a set of maps.
What am I missing?

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I think that the definition of $K(a,x,b,y)$ was typoed and should read as follows: $$K(a,x,b,y) = \{p \in P:p_{ax+by}=ap_x+bp_y\}.$$ This really is a closed set in the product: it’s the set of all points whose $(ax+by)$-th coordinate is a particular linear combination of their $x$-th and $y$-th coordinates. In a more familiar finite-dimensional setting, it’s analogous to $$\{(x_1,x_2,x_3,x_4,x_5)\in\mathbb{R}^5:x_4 = 2x_1 + 3x_5\},$$ for instance. Any point in $P$ that doesn’t satisfy the equation $p_{ax+by}=ap_x+bp_y$ can easily be bounded away from $K(a,x,b,y)$ by a basic open set in the product that restricts only the $x$-th, $y$-th, and $(ax+by)$-th coordinates, so $K(a,x,b,y)$ is closed in $P$.

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Thanks! That clarified a lot. But now I'm curious why the embedding $\iota(B)$ is truly set of all "linear" elements $p\in P$ ? Does this follow from construction of the embedding or..? –  Dustin Tran Oct 4 '11 at 4:25
    
@Dustin: Yes. Each $p\in P$ is a function from the vector space $\ell_1$ to a certain product of symmetric closed intervals. I’ve been writing $p_x$ for the $x$-th coordinate, but it may be clearer if I write $p(x)$. Then $\iota [B]$ is the set of $p\in P$ such that $p(ax+by)=ap(x)+bp(y)$ for all $a,b\in\mathbb{R}$ and $x,y\in\ell_1$. And you do get all of them: Robert showed how to reconstruct $y$ from $p=\iota(y)$. –  Brian M. Scott Oct 4 '11 at 4:39
    
Ah, I see now! What confused me previously was when to consider $p$ as maps when to consider $p$ as elements. –  Dustin Tran Oct 4 '11 at 4:44

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