State each of the following English sentences in symbols, as its final form, not containing $\lnot$.

(a): $f$ is not continuous at $a$

(b): The sequence $(a_n)_{n \in \mathbb{N}}$ approaches to $\infty$ as $n$ increases.

(c): $\lnot$(b)

Since i know that the formal definition of $f$ being continuos at $a$ is

$\forall \epsilon (\epsilon>0 \rightarrow \exists \delta (\delta>0 \land \forall x (|x-a|< \delta \rightarrow |f(x)-f(a)|<\epsilon)))$

My try is negating the whole statement by: $$\exists\epsilon(\epsilon>0 \land \forall\delta(\delta>0 \rightarrow \exists x(|x-a|<\delta \land |f(x)-f(a)| ))\geq\epsilon)$$ I'm not confident that this is equivalent to $f$ being discontinuous, because I can't come up with a graphical intuition it.

Likewise, if (b) is $$\forall M (M>0 \rightarrow \exists N_0 (N_0 \in \mathbb N \land (\forall n ((n \in N \land n>n_0)\rightarrow f(x)>M ))))$$

is $$\lnot(b) = \exists M(M>0 \land \forall N_0(N_0 \in N \rightarrow (\exists n ((n \in \mathbb N \land n>N_0)\land f(x)\le M))))$$ ?

  • $\begingroup$ What is the question? $\endgroup$ Mar 31 '20 at 6:50
  • $\begingroup$ @Taroccoesbrocco My question is whether i'm right or not. $\endgroup$
    – John. P
    Mar 31 '20 at 6:54
  • $\begingroup$ I added the "solution verification" tag for you. $\endgroup$ Mar 31 '20 at 11:50
  • $\begingroup$ In the definition of convergence to $\infty$, there is no need to require $M > 0$. It doesn't hurt anything, but it makes no difference if you require it or not. The exact same sequences will satisfy the condition with it as satisfy the conditions without it. This would also simplify the negation. $\endgroup$ Mar 31 '20 at 12:01

Both of your solutions are correct, though the second would be better stated without the unnecessary restriction on $M$, as indicated in my comment.

Concerning the "graphical intuition", the definition of continuity can be rephrased as "for every neighborhood of $f(a)$, there is a neighborhood of $a$ which is carried by $f$ into the neighborhood of $f(a)$. In other words: if you draw a circle about $f(a)$, then you can also draw a circle about $a$ such that everything inside it is carried into the circle about $f(a)$.

Your negation says there is some neigbhorhood of $f(a)$ such that any neighborhood of $a$ will contain at least one point that is carried outside the neighborhood of $f(a)$. This is what we mean by discontinuity: no matter how close you get to $a$, something is carried away from $f(a)$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.