$$
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\newcommand{\paren}[1]{\left(#1\right)}
\newcommand{\sq}[1]{\left[#1\right]}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\ang}[1]{\left\langle#1\right\rangle}
\newcommand{\floor}[1]{\left\lfloor#1\right\rfloor}
\newcommand{\ceil}[1]{\left\lceil#1\right\rceil}
\newcommand{\C}{\mathbb{C}}
\newcommand{\D}{\mathbb{D}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\T}{\mathbb{T}}
\renewcommand{\S}{\mathbb{S}}
\newcommand{\intr}{{\large\circ}}
\newcommand{\limni}[1][n]{\lim_{#1\to\infty}}
\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}
$$
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Office hours:
- Mondays 9:00-10:00
- Wednesdays 14:00-16:00
GSI:
Rahul Dalal
- M 10:30-12:30
- TTh 17:30-19:30
- WF 11:00-13:00
-
Suppose that $(S, \preceq)$ is an ordered set, that $E \subseteq S$, and that $t$ is a lower bound for $E$.
Then for any $x \prec t$, $x$ is a lower bound for $E$.
- True.
- True, but only with the additional assumption that $t = \inf E$.
- False.
-
Suppose that $(S, \preceq)$ is an ordered set, and $E \subseteq S$ is bounded above.
Then $E$ has infinitely many distinct upper bounds...
- ...with no further assumptions.
- ...provided that $E$ does not have a maximum element.
- ...provided that $S$ does not have a maximum element.
- ...if $S$ has the least upper bound property.
-
Suppose $(S, \square)$ is an ordered set, and $E \subseteq S$.
We say $E$ is bounded above if...
- ...for every $t \in E$ there is some $s \in S$ with $t \square s$.
- ...$\sup E$ exists.
- ...there is some $s \in S$ so that for every $t \in E$, $t \square s$.
- ...whenever $s \in S$ and $t \in E$, we have $t \square s$.
