Topology of pointwise convergence

**davismj** · December 18th 2010, 07:05 PM

Not really sure how to begin this one. Any help appreciated!

**Drexel28** · December 18th 2010, 11:09 PM

Originally Posted by davismj

Not really sure how to begin this one. Any help appreciated!

What do you mean topologized by the family? Give $\mathbb{X}$ the weak topology generated by the family of maps?

**davismj** · December 19th 2010, 06:53 AM

I have no idea what that means. I assume its the intersection of all the open balls of radius r given by the family of semi norms. There's a lot about this concept out there, especially with respect to locally convex spaces, but I still don't really understand it.

**Drexel28** · December 19th 2010, 10:23 AM

Originally Posted by davismj

I have no idea what that means. I assume its the intersection of all the open balls of radius r given by the family of semi norms. There's a lot about this concept out there, especially with respect to locally convex spaces, but I still don't really understand it.

So you're saying give it the metric topology where an open ball is $\displaystyle B\left(x_0;r\right)=\bigcap_{x\in[0,1]}\left\{y\in\mathbb{X}:\rho_x(y-x_0)<r\right\}$ ??

**davismj** · December 19th 2010, 10:34 AM

I'm guessing, although I believe that topology separates points, and I think the idea of the proof is to demonstrate te consequences of a non metrizable topological space, or something.

**Opalg** · December 19th 2010, 11:36 AM

Originally Posted by Drexel28

So you're saying give it the metric topology where an open ball is $\displaystyle B\left(x_0;r\right)=\bigcap_{x\in[0,1]}\left\{y\in\mathbb{X}:\rho_x(y-x_0)<r\right\}$ ??

It's not a metric topology, basically because there are uncountably many of the defining seminorms $\rho_x$ . A basic family of neighbourhoods of f would consist of sets of the form $\{g\in\mathbb{X}:\rho_{x_j}(f-g) = |f(x_j)-g(x_j)|<\varepsilon\; (1\leqslant j\leqslant n)\}$ for each finite set of points $x_1,...,x_n\in[0,1].$ It is called the topology of pointwise convergence because a sequence (or more generally a directed net) $\{f_i\}$ converges to the limit $f$ iff $f_i(x)\to f(x)$ for each point $x\in[0,1].$

For the last part of the question, I think that you have to use a cardinality argument. You want to find a sequence $(f_n)$ in $\mathbb{X}$ which converges to 0 (in the topology of pointwise convergence) so slowly that if you multiply its terms by any sequence of scalars that tends to infinity then the resulting sequence $(\gamma_nf_n)$ no longer converges to 0.

The set $\Gamma$ of all sequences of integers that tend to infinity has cardinality $2^{\aleph_0}$ , which is the same as the cardinality of [0,1]. So there exists a bijective map $\phi:\Gamma\to [0,1]$ . Define the function $f_n\in\mathbb{X}$ by $f_n(\phi(\gamma)) = 1/\gamma_n\ (\gamma\in\Gamma)$ .

Then $f_n(x)\stackrel{n}{\to}0$ for all $x\in[0,1]$ , but $\gamma_nf_n(\phi(\gamma)) = 1\stackrel{n}{\not\to}0.$

Finally, I leave you to check that if $(\gamma_nf_n)$ fails to converge to 0 for all sequences of integers tending to infinity, then the same will be true for all sequences of reals tending to infinity.

**Drexel28** · December 19th 2010, 11:57 AM

Originally Posted by Opalg

It's not a metric topology, basically because there are uncountably many of the defining seminorms $\rho_x$ . A basic family of neighbourhoods of f would consist of sets of the form $\{g\in\mathbb{X}:\rho_{x_j}(f-g) = |f(x_j)-g(x_j)|<\varepsilon\; (1\leqslant j\leqslant n)\}$ for each finite set of points $x_1,...,x_n\in[0,1].$ It is called the topology of pointwise convergence because a sequence (or more generally a directed net) $\{f_i\}$ converges to the limit $f$ iff $f_i(x)\to f(x)$ for each point $x\in[0,1].$

For the last part of the question, I think that you have to use a cardinality argument. You want to find a sequence $(f_n)$ in $\mathbb{X}$ which converges to 0 (in the topology of pointwise convergence) so slowly that if you multiply its terms by any sequence of scalars that tends to infinity then the resulting sequence $(\gamma_nf_n)$ no longer converges to 0.

The set $\Gamma$ of all sequences of integers that tend to infinity has cardinality $2^{\aleph_0}$ , which is the same as the cardinality of [0,1]. So there exists a bijective map $\phi:\Gamma\to [0,1]$ . Define the function $f_n\in\mathbb{X}$ by $f_n(\phi(\gamma)) = 1/\gamma_n\ (\gamma\in\Gamma)$ .

Then $f_n(x)\stackrel{n}{\to}0$ for all $x\in[0,1]$ , but $\gamma_nf_n(\phi(\gamma)) = 1\stackrel{n}{\not\to}0.$

Finally, I leave you to check that if $(\gamma_nf_n)$ fails to converge to 0 for all sequences of integers tending to infinity, then the same will be true for all sequences of reals tending to infinity.

I understand, that was why I was asking this since the OP made it seems like he thought it was a metric topology. I'd like to mention that this topology is homeomorphic to $[0,1]^{\mathbb{C}}$ with the product topology.

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