$$ \newcommand{\cis}{\operatorname{cis}} \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\paren}[1]{\left(#1\right)} \newcommand{\sq}[1]{\left[#1\right]} \newcommand{\abs}[1]{\left\lvert#1\right\rvert} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\ang}[1]{\left\langle#1\right\rangle} \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\C}{\mathbb{C}} \newcommand{\D}{\mathbb{D}} \newcommand{\R}{\mathbb{R}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} \newcommand{\F}{\mathbb{F}} \newcommand{\T}{\mathbb{T}} \renewcommand{\S}{\mathbb{S}} \newcommand{\intr}{{\large\circ}} \newcommand{\limni}[1][n]{\lim_{#1\to\infty}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cC}{\mathcal{C}} \newcommand{\cD}{\mathcal{D}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cG}{\mathcal{G}} \newcommand{\cH}{\mathcal{H}} \newcommand{\cI}{\mathcal{I}} \newcommand{\cJ}{\mathcal{J}} \newcommand{\cK}{\mathcal{K}} \newcommand{\cL}{\mathcal{L}} \newcommand{\cM}{\mathcal{M}} \newcommand{\cN}{\mathcal{N}} \newcommand{\cO}{\mathcal{O}} \newcommand{\cP}{\mathcal{P}} \newcommand{\cQ}{\mathcal{Q}} \newcommand{\cR}{\mathcal{R}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cT}{\mathcal{T}} \newcommand{\cU}{\mathcal{U}} \newcommand{\cV}{\mathcal{V}} \newcommand{\cW}{\mathcal{W}} \newcommand{\cX}{\mathcal{X}} \newcommand{\cY}{\mathcal{Y}} \newcommand{\cZ}{\mathcal{Z}} \newcommand{\bA}{\mathbb{A}} \newcommand{\bB}{\mathbb{B}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bD}{\mathbb{D}} \newcommand{\bE}{\mathbb{E}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bG}{\mathbb{G}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bI}{\mathbb{I}} \newcommand{\bJ}{\mathbb{J}} \newcommand{\bK}{\mathbb{K}} \newcommand{\bL}{\mathbb{L}} \newcommand{\bM}{\mathbb{M}} \newcommand{\bN}{\mathbb{N}} \newcommand{\bO}{\mathbb{O}} \newcommand{\bP}{\mathbb{P}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bS}{\mathbb{S}} \newcommand{\bT}{\mathbb{T}} \newcommand{\bU}{\mathbb{U}} \newcommand{\bV}{\mathbb{V}} \newcommand{\bW}{\mathbb{W}} \newcommand{\bX}{\mathbb{X}} \newcommand{\bY}{\mathbb{Y}} \newcommand{\bZ}{\mathbb{Z}} $$

Useful links

Office hours:

  • Mondays 18:00-19:00
  • Wednesdays 14:00-16:00

Email

GSI:

Nima Moini
  • Mondays 10:00-14:00
  • Tuesdays 10:00-14:00
  • Wednesdays 10:00-12:00

Exams

Let $(\mathcal{M}, d)$ be a metric space.

  1. If every sequence in $\mathcal{M}$ has a convergent subsequence, then $\mathcal{M}$ is complete.
    1. True
    2. False
  2. If $\mathcal M$ is complete, then every sequence in $\mathcal M$ has a convergent subsequence.
    1. True
    2. False

  3. Suppose $(x_n)_n$ is a sequence in $\R$, and $X = \set{x_n \mid n \in \N}$. We wish to prove that if $x \in X'$ there is a subsequence converging to $x$. Is the following proof valid?

    For each $k \in \N$, since $x$ is a limit point of $X$, there is some $y \in X$ so that $d(x, y) \lt \frac1k$. Since $y$ occurs in the sequence, there is some index $n_k$ so that $x_{n_k} = y$. Now if we take $(x_{n_k})_k$, this converges to $x$ by construction (indeed, given $\epsilon \gt 0$, if $k \gt \frac1\epsilon$ we have $d(x_{n_k}, x) \lt \frac1k \lt \epsilon$). Thus there is a subsequence converging to $x$.

    1. Yes
    2. No
  4. Suppose $F \subseteq \mathcal M$ and $(x_n)_n$ is a sequence in $F$ which converges. Then the limit of the sequence is a limit point of $F$.
    1. True
    2. True as long as $F' \neq \emptyset$
    3. False
  5. Are these polls useful/helpful/worth doing?
    1. Yes
    2. No
    3. Meh