soft question - How to stop forgetting proofs - for a first course in Real Analysis?

I am taking my first course in analysis. I like the subject. I study it almost on a daily basis. I try to prove theorems on my own without even looking at the hints. If I really get stuck I just read the first line of the proof and then try to continue on my own. I find this approach to be very rewarding. The problem is that I tend to forget all the proofs later. Some I still remember because I liked the idea of the proof, but most of the others I will forget fast. What am I doing wrong? Any tips on how to study analysis?...Read more

real analysis - Weak topology generated by family of functions is weakest wrt which each of the given functions is continuous

Let $\mathcal{F}=\{ f_\alpha:X \rightarrow X_\alpha, \alpha \in A\}$ be a family of functions from X onto topological spaces $X_\alpha$. The weak topology for X generated by functions $f_\alpha$ is the topology determined by the subbasis of open sets $\{ f^{-1}_\alpha (O_\alpha): \alpha \in \mathcal{A}, O_\alpha$ open in $X_\alpha\}$.Why would the weak topology defined above be the weakest topology with respect to each of which of the given function is continuous?Suppose we have a family of constant functions (they are continuous), then the top...Read more

real analysis - Choice of $q$ in Baby Rudin's Example 1.1

First, my apologies if this has already been asked/answered. I wasn't able to find this question via search.My question comes from Rudin's "Princicples of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2. In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof. As an exercise, I tried to d...Read more

real analysis - Prove that every subset of $\mathbb{N}$ is clopen

My proof is:Let $x$ be any point in $\mathbb{N}$, and let $\beta$ be the closest number (such as a factor or irrational number) to $x$, such that $\beta \notin \mathbb{N}$ and $\beta > x$.Let $B_\epsilon(x)$ be the epsilon ball around $x$ in $\mathbb{N}$. Let $\epsilon = \beta - x$.Clearly $\forall x \in \mathbb{N}: \exists \epsilon >0, \epsilon \in \mathbb{R}: B_\epsilon \left({x}\right) \subseteq \mathbb{N}$. So we have that all singleton sets in $\mathbb{N}$ are open.All subsets of $\mathbb{N}$, except the empty set, consist of a union...Read more

real analysis - Existence of a subsequence in compact metric space

Let $\{x_n\}$ be a sequence in a compact metric space X such that: 1) For all $n \in \mathbb{N}$ the set $\{ x_n : n \ge N\}$ is closed. 2) The sequence $\{x_n\}$ is Cauchy. Prove that there exists $x \in X$ and a subsequence $\{x_{n_k}\}$ such that $x_{n_k} =x$ for all $k \in \mathbb{N}$.Since we are in a compact metric space, the Cauchy sequence is convergent to some $x \in X$ and as $\{x_n\}$ is closed that would mean that $x= x_M$ for some $m\in \mathbb{N}$. But I'm not sure how to extract a subsequence which is eventually constant....Read more

real analysis - I am stumped on this problem

It is from Intro to Analysis by Bartle 3rd edLet $I:=[a,b]$ let $f:I\rightarrow\mathbb{R}$ be a continuous function with $f(x)>0$ Prove that there exists a number $c>0$ such that $f(x)\geq c $ for all in $I$I am given 2 hints: Use Boundedness theorem with $1/f$ orMax-min theoremIf $f$ is bounded then $1/f$ is unbounded? Can this type of theorem be done without proof by contradiction?Mock proofIf $f$ is continuous then it is bound. So we have $|f(x)|<M$For any number $n$ in $\mathbb{N}$ there is a number$x_n$ in $I$ s.t $|f(x_n)|<n$...Read more

real analysis - Convergence of indicator functions series

Say $E$ is a measurable set, and $\{E_k\}$ is a series of measurable sets defined by$$E_k \subset E, m(E \setminus E_k) < \frac{1}{k}, k = 1, 2, 3, ...$$Do their corresponding indicator functions series converge to 1, almost surely?$$\lim_{k \to \infty} 1_{E_k} = 1, a.s. x \in E$$If $E_k \subset E_{k+1}$, then it would be trivial to prove the above statement with $A = \bigcap_{k=1}^{\infty} (E \setminus E_k)$. However, without this condition, I'm getting suspicious about its correctness, but I cannot raise any counterexample either....Read more

real analysis - Understanding Proof that there is no Onto function between set and its power set

I am having hard time understanding the proof that there is no onto function from set A to its power set. In the proof author has constructed the set B = {$a \in A : a \notin f(a)$}. I have taken few sets and constructed functions but i am not able to get the intuition behind construction of set B , i mean to say how should we know that set B has to be constructed in this wayThanks...Read more

real analysis - Show that if a subset $E$ of a compact metric space $X$ is compact in $X$, then it is closed in $X$.

I am self-studying Royden's Real Analysis; Exercise 58 of Section 9.5, "Compact Metric Spaces", asks: Let $E$ be a subset of the compact metric space $X$. Show that the subspace $E$ is compact if and only if $E$ is a closed subset of $X$.The reverse implication is easy: if $E$ is closed, given an open cover $\mathcal{O}$ of $E$, $\mathcal{O} \cup \{X \setminus E\}$ is an open cover of $X$, which, by compactness of $X$, contains a finite subcover $\mathcal{O}'$ of $X$; removing $X \setminus E$ from $\mathcal{O}'$ if necessary, we arrive at a fi...Read more

real analysis - A metrisable compact space that is non isometric to any compact subset of $\mathbb{R}^{n}$

I'm looking for the metrisable compact $M$ so that there is no isometry $f:M \rightarrow \mathbb{R}^{n}$ with $im(f) = K$ $-$ compact subspace.First, isometry preserves completeness, separability, and boundedness. Then, let's see how do the compact subspaces of $\mathbb{R}^{n}$ look like. First, they are complete (this works not only in the case of $\mathbb{R}^{n}$ but is the cases of metrisable space), then they are bounded (Heine-Borel states that compact set is bounded and closed). Moreover, any compact space in metric space is separable. So...Read more