# Thread: Topology of pointwise convergence

1. ## Topology of pointwise convergence

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

2. 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?

3. 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.

4. 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)??

5. 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.

6. 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)??
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.

7. 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.