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.

Thing	Median	Mean	Standard Deviation
Assignment 1	30/50	~29.4/50	~15.1/50
Assignment 2	28/50	~26.5/50	~15.5/50
Assignment 3	28/50	~26.4/50	~15.0/50
Assignment 4	30/50	~31.1/50	~16.3/50
Assignment 5	32/50	~27.5/50	~16.4/50
Midterm 1	37/60	~33.4/60	~14.7/60

Remember that your grade on the final exam will replace your grade on one of the midterms if doing so is to your benefit.