Ian Charlesworth

Math 104 - Introduction to Analysis

Midterm 1 post-mortem

Some comments about common mistakes and issues from grading the midterm:

• Knowing the definitions is essential. The order of quantifiers in definitions is not open to interpretation. In particular, every symbol used in a definition should be quantified, and the order in which they are quantified is important. Statements like "There is a $N \in \N$ so that if $n \gt N$ then $d(a_n, L) \lt \epsilon$ for every $\epsilon \gt 0$" are ambiguous: does this mean that there is one $N$ that works for every $\epsilon$, or for each choice of $\epsilon \gt 0$ there is some choice of $N$?

• Especially in 2B, there were a lot of people beginning their proofs with "Since $x \gt 1$ and $y \gt 2x \gt 2$ we have $x + y \gt 3$." There is a subtle problem here which is that there is no quantification on $x$ or $y$. In the definition of the set \[\set{x+y \mid x \gt 1, y \gt 2x}\] both $x$ and $y$ only make sense within the definition of the set in question; by the next line, they've gone out of scope and we've forgotten what they are. For example, suppose there were two sets defined: \begin{align*}S_1 &= \set{x+y \mid x \gt 1, y \gt 2x} \\ S_2 &= \set{x + y \mid x \lt 0, y \lt 1}.\end{align*} Is it true that "because $x \gt 1$, we cannot have $x \lt 0$, so $S_2$ is empty"? No, because the restrictions on $x$ only having meaning within the definition of the set $S_1$ or $S_2$; they don't reach outside.

  A better argument would begin "If $x \gt 1$ and $y \gt 2x$ then $y \gt 2$ so $x + y \gt 3$, and since any $z \in S_1$ is of this form we have $z \gt 3$, i.e., 3 is a lower bound for $S_1$..."

• If you are going to cite a theorem, clearly state its hypotheses and conclusion, or its name, or both. There were a few people who wrote things like "By a theorem, this is true" when often the "this" in question was not true.
• Related to the above, if you are going to cite a theorem, make sure you know its hypotheses and conclusion. In particular, the Heine-Borel theorem only holds in the metric spaces $(\R^n, d_n)$ where $d_n$ is the usual distance metric. In the metric space $(\R, d_{discrete})$, the set $\set{x \in \R \mid 0 \lt x \lt 1}$ is closed and bounded but not compact. In fact, any infinite set is closed and bounded but not compact.

Here is some data about how grades are shaping up so far.