Negation of a statement regarding a closed set

Question

I am trying to negate the following statement:

If $(x_n)$ is an arbitrary sequence, $x_n \in K$ (where $K$ is a set) $\forall n>0$ and $\lim \limits_{n \to \infty}x_n=x$ with $x \in \mathbb{R}$ then $x \in K$

I believe the answer is either:

If $\exists$ a sequence $ (x_i)$ with $x_i \in K$ (where $K$ is a set) $\forall i>0$ and $\lim \limits_{i \to \infty}x_i=x$ with $x \in \mathbb{R}$ then $x \notin K$

or:

If $\exists$ a sequence $ (x_i)$ with $x_i \in K$ (where $K$ is a set) for some $i>0$ and $\lim \limits_{i \to \infty}x_i=x$ with $x \in \mathbb{R}$ then $x \notin K$

Would someone be able to explain logically (not using rules) why one is correct over the other?

Answer 1

1. The statement is true. Please check the definition of a closed set.

2. Changing finitely many terms of a sequence does not influence its limit, i.e. the limit does not depend on the first $n$ terms for any fixed $n$. Intuitively, the limit is something the sequence tends to, and changing the first few steps do not influence what it tends to.

Edit: Given your edited question, the statement could be disproved by a counter example, e.g $K=(0,1)$, $x_n=\frac{1}{n}$.

If $K$ is closed the statement is true. Otherwise both can happen. Of course our discussion is in the topology of $\mathbb{R}$

Edit_2: It seems that I have drastically misunderstood your question. If you only want the negation of that statement, it should be: There exists a sequence $(x_n)\subset K$ converging in $\mathbb{R}$ to $x$ such that $x\notin K$.

This condition does imply that $K$ is not closed, since $x$ is a limit point of $K$ but is not in $K$.

If we analyzed your answer, we could reformulate it as: if there exists a sequence $(x_n)\subset K$ converging in $\mathbb{R}$ to $x$, then $x\notin K$. I don't think this makes much sense...

I believe your confusion is with the quantifier? Your initial statement is of the form: FOR ANY "thing" that satisfies a certain property $P$, this "thing" satisfies another property $Q$. So the negation should be of the form: THERE EXISTS a thing that satisfies property $P$, but does not satisfy property $Q$

Answer 2

Your premise is wrong. A closed set does in fact contain all of its limit points. The statement you are trying to negate is actually true. Instead, try to convince yourself that if $K \subseteq \mathbb{R}$ is closed, and $\{x_n\}$ is a sequence of points in $K$ converging to $x \in \mathbb{R}$ then $x \in K$.

One additional note, we tend to write $\lim_{n\to \infty} x_n = x$ or $x_n \to x$, but we never write $\lim_{x_n \to \infty} = x$.

Answer 3

The negation of a statement of the form "if $ A $ then $ B $" is "$ A $ and not $ B $".

In your statement,

$ A $ is "$(x_n)$ is an arbitrary sequence, $x_n \in K$ (where $K$ is a set) $\forall n>0$ and $\lim \limits_{n \to \infty}x_n=x$ with $x \in \mathbb{R}$".

$ B $ is "$ x \in K $".

Therefore not $ B $ is "$ x \notin K $".

Hence the negation of your statement is "$(x_n)$ is an arbitrary sequence, $x_n \in K$ (where $K$ is a set) $\forall n>0$ and $\lim \limits_{n \to \infty}x_n=x$ with $x \in \mathbb{R}$ and $x \notin K$".

In other word, the negation(for your statement) is

for each sequence $(x_n)$ of $K$, the limit of $(x_n)$ belongs to $\mathbb{R}$ but it does not belong to $K$.

But, if $K$ is a closed set then for each sequence $(x_n)$ of $K$, the limit of $(x_n)$ is also belong to $K$.

Therefore if then there exists a sequence $(x_n)$ of $K$ such that the limit of $(x_n)$ does not belong to $K$ then $K$ is not closed. This is called the contraposition of the above statement.