# [Notes](#fn1): Problem Class, 12 Oct 2020
> Group 11 -- Problem Class, MATH40001/40009 IUM
> Stergios M. Antonakoudis ([e-mail])

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Welcome to IUM Problem Class Group 11. \
Today's class: <u>**Problem Sheet 2 of Part I.**</u>

[^1]: <small> *These notes[^1] are typed up and created during and for the purposes of the discussion in the online session of Problem Class Group 11 for IUM on MS Teams, similarly to the use of a 'whiteboard' in an actual class.* </small>

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## A brief overview: ==sets==

* $\quad A\subset B \qquad$--- *means*
$\quad$ If $x \in A$, then $x \in B$.

* $\quad A=B \qquad$--- *means*
$\quad$ If $x \in A$, then $x \in B$ $\quad (A\subset B)$,
$\qquad \forall x \in A,\quad x \in B$ *(as propositions)*, and
$\quad$ If $x \in B$, then $x \in A$ $\quad (B\subset A)$,
$\qquad \forall x \in B,\quad x \in A$ *(as propositions)*.

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## Problems to discuss: ==1, 2, 3, 4, 5==

### Problem 1
If $X,Y$ are sets, prove that $X\cap Y = Y \cap X$.
### Solution 1 to Problem 1

First, we need to show that $X\cap Y \subset Y \cap X$:

Let $x \in X \cap Y$. <u>Goal</u>: we need to show $x \in Y \cap X$.

Hence, $x \in X$ and $x \in Y$, but this is the same as $x \in Y$ and $x \in X$. In other words -- *by definion* -- $x \in Y \cap X$. Which is what we wanted to show.

Second, we need to show that $Y\cap X \subset X \cap Y$:

Let $x \in Y \cap X$. <u>Goal</u>: we need to show $x \in X \cap Y$.

Hence, $x \in Y$ and $x \in X$, but this is the same as $x \in X$ and $x \in Y$. In other words -- *by definion* -- $x \in X \cap Y$. Which is what we wanted to show.

QED

### Sketch Solution 2 to Problem 1

*Informally first*: If you'd like to prove that a set $A$ is a subset of $X \cap Y$ i.e. $A \subset X \cap Y$. You need to prove/show that $A \subset X$ and $A \subset Y$.

<u>1st GOAL</u>: $X\cap Y \subset Y$ and $X\cap Y \subset X$ (i.e. $X\cap Y \subset Y \cap X$).

(spelling out the formal proof of the 1st goal in detail left as exercise)

<u>2nd GOAL</u>: $Y\cap X \subset X$ and $Y\cap X \subset Y$ (i.e. $Y\cap X \subset X \cap Y$).

(spelling out the formal proof of the 1st goal in detail left as exercise)

QED

### Problem 5
(a) $\exists x \in \emptyset, 2+2=5$
(b) $\forall x \in \emptyset, 2+2=5$

### Solution to Problem 5
(a) is **_False_**.

> *~~Informally~~*, any proposition that starts with $\exists x \in \emptyset, \ldots$ is **FALSE**.

Since, $\exists x \in \emptyset$ is **FALSE**,
because there is no element $x$ so that $x \in \emptyset$ is true.

> Recall, $(False) \land P$ is always **False**.

(b) is **_True_**.

> *~~Informally~~*, any proposition that starts with $\forall x \in \emptyset, \ldots$ is **TRUE**.

Since, $\forall x \in \emptyset$ is **FALSE**, because there is no element $x$ so that $x \in \emptyset$ is true.

> Recall, $(False) \implies P$ is always **True**.
> equivalenly, $(True) \lor P$ is always **True**.

<u>*Interlude*</u>:
* $\lnot (\, \forall x, P(x)\, ) \iff \exists x, \lnot P(x)$
* $\forall x, P(x) \iff \lnot (\, \exists x, \lnot P(x) \, )$

### Problem 4
(a) $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, x+y = 2$.
(b) $\exists y \in \mathbb{R}, \forall x \in \mathbb{R}, x+y = 2$.

### (Sketch) Solution to Problem 4
(a) Let $P(x)$ denote $\exists y \in \mathbb{R}, x+y = 2$.
The proposition we're given is $\forall x \in \mathbb{R}, P(x)$.
It's logical negation is $\lnot (\, \forall x \in \mathbb{R}, P(x)\, )$,
which is equivalent to $\exists x \in \mathbb{R}, \lnot P(x)$, but
$x \in \mathbb{R}$ and $\lnot P(x)$ cannot both be true (exercise!). Hence, $\exists x \in \mathbb{R}, \lnot P(x)$ is **False**; and therefore, $\forall x \in \mathbb{R}, P(x)$ is **True**.

(b) Skip for now.

### Problem 3

If $f:\mathbb{R} \rightarrow \mathbb{R}$ is a function, then, we say that $f$ is continuous at $x \in \mathbb{R}$ if
$\forall \epsilon \in \mathbb{R}_{>0}$, $\exists \delta \in \mathbb{R}_{>0} \forall y \in \mathbb{R}, |y-x| < \delta, |f(y)−f(x)|< \epsilon$.
It's negations is

$\exists \epsilon \in \mathbb{R}_{>0}, \forall \delta \in \mathbb{R}_{>0}, \exists y \in \mathbb{R}, |y-x| < \delta, |f(y)-f(x)| \geq \epsilon$.

Compare with:

$\forall \epsilon \in \mathbb{R}_{>0}$, $\exists \delta \in \mathbb{R}_{>0} \forall y \in \{y \in \mathbb{R}, |y-x| < \delta \}, |f(y)−f(x)|< \epsilon$.
It's negation is

$\exists \epsilon \in \mathbb{R}_{>0}, \forall \delta \in \mathbb{R}_{>0}, \exists y \in \{y \in \mathbb{R}, |y-x| < \delta\}, |f(y)-f(x)| \geq \epsilon$.