$$ \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

  1. Suppose that $X, Y, Z$ are metric spaces, and $f: X \to Y$ and $g : Y \to Z$ are so that $g$ is continuous and $g \circ f$ is continuous. Then $f$ must be continuous.
    1. True.
    2. False.
  2. Suppose that $X, Y$ are metric spaces and $f : X \to Y$ has the property that every open set $U \subseteq X$ has open image $f(U) \subseteq Y$. Then...
    1. ...$f$ is continuous.
    2. ...$f$ is continuous, provided that $X$ is compact.
    3. ...$f$ is continuous, provided that both $X$ and $Y$ are compact.
    4. ...$f$ need not be continuous, even if $X$ and $Y$ are compact.
  3. Suppose that $X, Y$ are metric spaces and $f : X \to Y$ has the property that every closed set $F \subseteq Y$ has closed preimage $f^{-1}(F) \subseteq X$. Then...
    1. ...$f$ is continuous.
    2. ...$f$ is continuous, provided that $X$ is compact.
    3. ...$f$ is continuous, provided that both $X$ and $Y$ are compact.
    4. ...$f$ need not be continuous, even if $X$ and $Y$ are compact.
  4. Suppose that $X, Y$ are metric spaces and $f : X \to Y$ is continuous. The $f$ is uniformly continuous, provided...
    1. ...$X$ and $Y$ are compact.
    2. ...$X$ is compact.
    3. ...$Y$ is compact.
    4. ...at least one of $X$ and $Y$ is compact.

  5. We want to prove that if a subset $E$ of a metric space $X$ is connected, then so is $\overline E$. Is the following proof valid?

    Suppose that $U_1, U_2$ are a disconnection of $\overline E$. Then since $\overline E \cap U_1 \neq \emptyset$, it must be the case that $E \cap U_1$ is non-empty (for otherwise $U_1^c$ is a closed set containing $E$, and so $\overline{E} \subseteq U_1^c$). Likewise, $E \cap U_2$ is non-empty. Since $E \subseteq \overline E \subseteq U_1\cup U_2$, and by assumption $U_1 \cap U_2 = \emptyset$, these give a disconnection of $E$.

    1. Yes
    2. No