2
$\begingroup$

By logic formula of pointwise convergence: $$(\forall \varepsilon>0) (\forall x\in S) (\exists n_0) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon)$$

It can be interpreted by set theroy, for example: $$\cup_{m}\cap_{n} \cap_{n_0(m,n)}E_{n,m}$$ where $E_{n,m}={\{x~|~~|f_n(x)-f(x)|<1/m}\}$

However, by the logic formula of uniform convergence: $$(\forall \varepsilon>0) (\exists n_0) (\forall x\in S) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon)$$

the set is (if I am not wrong): $$\cup_{m}\cap_{n_0(m)}E_{n,m}$$

I know the order of $(\forall n_0)(\exists x), $ $(\exists x)(\forall n_0)$ makes the uniform and pointwise convergence differently. However, as in the example of pointwise convergence, the formula can be translated one by one literally when taken the term ($\forall x)$ and translate the formula mechanically, however, it is not in the uniform case and need a second thought.

So my question is why the insetsection of $\cap_{n_0}$ missing from the formula. I am confused that how to translate the (proper that can be translated) logic formula into exact set unions and intersections mechanically and correctly, especially in the occasions of the real in analysis / measure theory.

$\endgroup$
2
$\begingroup$

You're rendering these incorrectly. $\forall, \exists$ correspond to $\bigcap, \bigcup$ respectively. This should be clear from thinking about what the symbols mean.

For pointwise convergence, the set of $x$ such that $f_n(x)\to f(x)$ pointwise is $$ \bigcap_{m\in \Bbb N} \bigcup_{n_0\in \Bbb N}\bigcap_{n>n_0} E_{n,m} $$ For uniform convergence, there really isn't a corresponding set: it doesn't make sense to speak of "the $x$ at which $f_n(x)\to f(x)$ uniformly", because uniform convergence is a property of sets of $x$s and not of individual points $x$.

How, in words, would you describe the set you're trying to define for uniformly convergent $f_n\to f$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.