- Thread starter
- #1

- Thread starter dray
- Start date

- Thread starter
- #1

- Admin
- #2

- Jan 26, 2012

- 4,183

$A$ is dense in $\bar{A}$, by definition, if and only if any point $\bar{a}\in\bar{A}$ belongs either to $A$ or is a limit point of $A$. Is that true?I'm just working on a problem and wanted some clariciation.

Let $X$ be a metric space and $A$ a non-empty subset of $X$. Is it true that $A$ is dense in $\bar{A}$? Its just that I am assuming this in my proof.

- Thread starter
- #3

So I have started by assuming that $A$ is dense in $\bar{A}$ so that for any $x\in{\bar{A}}$ we can construct a sequence of points $(x_n)$ that converge to $x$, where each $x_n\in{A}$.

- Admin
- #4

- Jan 26, 2012

- 4,183

Well, I can appreciate that this is one step in a different proof, but you haven't answered my question in Post # 2. What is the closure of $A$?

So I have started by assuming that $A$ is dense in $\bar{A}$ so that for any $x\in{\bar{A}}$ we can construct a sequence of points $(x_n)$ that converge to $x$, where each $x_n\in{A}$.

- Thread starter
- #5

I don't know, since all we are told is that $A$ is a subset of a metric space $X$. Hence my initial question.Well, I can appreciate that this is one step in a different proof, but you haven't answered my question in Post # 2. What is the closure of $A$?

- Admin
- #6

- Jan 26, 2012

- 4,183

Here's the closure definition. With that information, can you answer the question in post # 2?I don't know, since all we are told is that $A$ is a subset of a metric space $X$. Hence my initial question.

- Thread starter
- #7

- Admin
- #8

- Jan 26, 2012

- 4,183

Great! So your answer is...Answer to your post would be "yes".

And that tells you that the answer to your original post is...

- Thread starter
- #9

- Admin
- #10

- Jan 26, 2012

- 4,183

Right. A slightly more standard notation would be curly braces: $\bar{A}=\{x\in{X}:d(x,A)=0\}.$Thanks.

I've just realised that $\bar{A}=({x\in{X}:d(x,A)=0})$, which would have answered my original question.

Brain fried!

- Thread starter
- #11

Ahhh....I needed to include \ before the {. I'm a $LaTeX$ novice.Right. A slightly more standard notation would be curly braces: $\bar{A}=\{x\in{X}:d(x,A)=0\}.$

- Admin
- #12

- Jan 26, 2012

- 4,183

Understood. Actually, you would write it as $\LaTeX$. The curly brace is the basic way to organize collections of thingsAhhh....I needed to include \ before the {. I'm a $LaTeX$ novice.

Cheerio.