# Thread: Show A is not a closed set, but the union of A and {0} is a closed set.

1. ## Show A is not a closed set, but the union of A and {0} is a closed set.

So basically, I have no idea how to answer this problem. I tried a few different approaches, including trying to show that the complement of A was closed. I am stuck at the very beginning as I do not know where to start with this problem. Could someone please, at the very least, outline what I am supposed to do to prove this? Thank you.

Here's the problem:

Show that $A= \{\frac{1}{n} : n\in N\}$ is not a closed set, but that $A \cup \{0\}$ is a closed set.

2. Originally Posted by seamstress
So basically, I have no idea how to answer this problem. I tried a few different approaches, including trying to show that the complement of A was closed. I am stuck at the very beginning as I do not know where to start with this problem. Could someone please, at the very least, outline what I am supposed to do to prove this? Thank you.

Here's the problem:

Show that $A= \{\frac{1}{n} : n\in N\}$ is not a closed set, but that $A \cup \{0\}$ is a closed set.
A set is closed if it contains all of its limit points. What are the limit point(s) of A?

3. Okay, I see what you are saying.
Right now, I have supposed c is a cluster point of A.
I used the definition of a cluster point to say that every e-neighborhood of c contains a point x in A such that x is not equal to c
==> x is contained in the e-neighborhood of c= (c-e, c+e) for some e>0
==> |x-c| < e

1. is this right?
2. How would I go about showing 0 is a cluster point and that A doesn't contain it? I'm slightly confused as what to do with the e.

thank you!

4. Please re-read my second post. I made a huge mistake while typing but just fixed it. what do you think?

5. 0 is a boundary point of A because every neighborhood of zero contains a point in A (namely 1/n for some n sufficiently large), and a point not in A (namely 0). So A is not closed.

6. Okay, I understand that. And I also understand that 0 is the only limit point of A, but I dont know how to prove that rigorously ?

7. Let $U$ be an $\epsilon$ neighborhood of $0$. So $x\in U \Leftrightarrow -\epsilon . Choose $n\in \mathbb{N}$ such that $n>\frac{1}{\epsilon}$. Then $\frac{1}{n}<\epsilon$, and we have found an element of $A$ in $U$. Thus $0$ is a limit point of $A$.

8. yes, but how would i prove that 0 is the only limit point of A?

9. If $x\ne \frac{1}{n}$ for all $n$, let $\epsilon = \inf \{ |x-\frac{1}{n}| : n\in \mathbb{N}\}$, and then argue that $\epsilon > 0$ and the $\epsilon$ neighborhood of $x$ does not contain any points of $A$.

Do the case $x=\frac{1}{n}$ separately.