Consider $ f : \mathbb{R} \to \mathbb{R} $ with $ f(x) = 0 $ at irrational $ x $ and $ f(\frac{p}{q}) = \frac{1}{q} $ at rationals. This is Thomae’s function. Btw when we say “a rational $ \frac{p}{q} $” it will be implicit that $ p, q $ are integers, $ q > 0 $ and the fraction is in reduced form. $ f $ is discontinuous at rationals. ... Read more 05 Aug 2021 - 1 minute read
$\color{goldenrod}{\text{Th}}$: Let $ (g(k)) $ be a sequence in a complete normed space, and $ (a_k) $ a sequence in $ \mathbb{R}_ {\geq 0} $ decreasing to $ 0 $. Suppose seq $ G(k) := g(1) + \ldots + g(k) $ is bounded. Then $ \displaystyle \sum_{k=1}^{\infty} a_k g(k) $ converges. $\color{goldenrod}{\text{Pf}} $: To get rid of $ g(k) $s, we ca... Read more 05 Aug 2021 - 2 minute read
Consider functions $ f_n (x) = \frac{1}{1+n^2 x^2} $. Pointwise limit $ \displaystyle f(x) = \lim_{n\to\infty} f_n (x) $ is $ 1 $ at $ 0 $ and $ 0 $ at every non-zero $ x $ (esp discontinuous). Plotting in geogebra shows the graphs of $ f_n $ getting ‘‘pinched”. Read more 04 Aug 2021 - less than 1 minute read
Here is one approach (another being this). Let $ (x_n) $ be a bounded sequence of reals. Thanks to Bolzano-Weierstrass, $ A := \lbrace \text{ accumulation points of }(x_n) \rbrace $ is non-empty. $ A $ is also bounded above, so $ a := \sup(A) $ exists, which we call $ \color{goldenrod}{\limsup x_n} $. $ a \in A $, i.e. $ a $ itself is an acc... Read more 04 Aug 2021 - 1 minute read
Let $ (x_n) $ be a sequence of reals. $ p \in \mathbb{R} $ is an $\color{goldenrod}{\text{accumulation point}}$ of the sequence if for every $ \epsilon > 0 $, $ x_n $ lies in $ (p-\epsilon, p+\epsilon) $ for infinitely many $ n $ (stricten “infinitely many” to “all but finitely many”, and we get definition of limit of a sequence). A subsequ... Read more 03 Aug 2021 - 1 minute read