0
$\begingroup$

I am trying to figure out if my solutions to my homework problems are correct. I also have a question about the process used to negate quantifiers. How do we know what to negate exactly? Sometimes it seems like we are not negating objects (like a complement of a set), but we are instead negating properties of those objects. However, sometimes it seems this is not true. For example, how come for part (b) when negating that statement we don't have something like 'there exists a subset of $Y \neq X$ that is open.' Could someone clarify this if possible?

What I did to solve these was to translate them into predicate logic then use things like deMorgan's laws and stuff like the negation of 'for all...' is 'there exists' and vice versa.

Problems: For each of the following statements write its negation and, for parts (a) and (b), state whether the negation is true or false. Need not provide proof, but you may give a justification.

(a) For every metric space $(X,d)$ there exist $p,q \in X$ such that $d(p,q) > 0.$

Negation: There exists a metric space $(X,d)$ such that for all $p,q \in X$ we have $d(p,q) \leq 0$. This is false by definition since a metric space requires that the metric be non-negative.

(b) There exists a metric space $(X,d)$ such that every subset of $X$ is closed.

Negation: For all metric spaces $(X,d)$ there exists a subset of $X$ that is open. This statement is false since not every metric space $X$ has a subset that is open. Take a subset of the natural numbers, for example.

(c) Let $f:[0,1] \rightarrow \mathbb{R}.$ For all $x_{0} \in [0,1]$ and all $\epsilon > 0$ there exists $\delta >0$ such that if $x \in [0,1]$ satisfies $|x-x_{0}| < \delta$ then $|f(x)-f(x_{0})| < \epsilon.$

Negation: There exists a $y \in \mathbb{R} \setminus [0,1]$ or there exists $\epsilon \leq 0$ such that for all $\delta \leq 0,$ $x\in [0,1]$ or $|x-x_{0}| < \delta$ or $|f(x)- f(x_{0})| \geq \epsilon.$

The homework doesn't ask us to say if it is false or not, and I'm not sure why, but if it did, would the answer be that it is false since there are contradictions in the statement. For example, $\epsilon, \delta$ begin less than or equal to zero doesn't make to much sense.

$$ $$ Logic translations:

(a) in logical symbols: $\forall (X,d) \hspace{2mm}\exists p,q \in X $ $( d(p,q) > 0).$

Negation: $\exists (X,d)$ $(\forall p,q \in X, d(p,q) \leq 0).$

(b) in logical symbols: $\exists (X,d) $ $(\forall E$ $ \subset X$, $ E = \bar{E})$

Negation: $\forall (X,d)$ $(\exists E \subset X$, $E = E^{o})$

$E^{o}$ is the set of all interior points of $E$ and $\bar{E}$ is the closure of $E$ -- used here just for notation.

(c) in logical symbols:

$\forall x_{0} \in [0,1] \wedge \forall \epsilon > 0$, $\exists \delta >0$ $((x \in [0,1] \wedge |x-x_{0}| < \delta) \rightarrow |f(x)-f(x_{0})| < \epsilon)$

Negation:

$\exists y \in \mathbb{R} \setminus [0,1] \vee \exists \epsilon \leq 0,$ $\forall \delta \leq 0$ $((x \in [0,1] \wedge |x-x_{0}| < \delta) \wedge |f(x) -f(x_{0})| \geq \epsilon)$.

If my logic translations are wrong, please point it out.

EDIT: I change my mind for the negation of part (b). Every set contains the empty set, which is open. So the negation is true.

$\endgroup$
  • 1
    $\begingroup$ "Non-negative" and "$d(p,q)\le0$" aren't mutually exclusive! $\endgroup$ – Kenny Lau Oct 1 '17 at 23:51
  • $\begingroup$ Not sure what you are implying... $\endgroup$ – physicsmajor Oct 1 '17 at 23:55
  • $\begingroup$ Look at your "justification" of (a) $\endgroup$ – Kenny Lau Oct 1 '17 at 23:56
  • $\begingroup$ You are saying that $d(p,q) \leq 0$ and $d$ being "non-negative" are not the same? $\endgroup$ – physicsmajor Oct 2 '17 at 0:01
  • 1
    $\begingroup$ For part (b), kindly note that the negation of "closed" is "not closed", not "open". $\endgroup$ – Kenny Lau Oct 2 '17 at 0:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.