Dense Subset in a Metric Space

[dray]

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.

 

[Ackbach]

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.

$A$ 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?

 

[dray]

Well I am trying to show that if $f:{X}\rightarrow{Y}$ and $g:{X}\rightarrow{Y}$ are continuous functions such that $f(x)=g(x)$ for all $x\in{A}$ then $f(x)=g(x)$ for every $x\in{\bar{A}}$.

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 $x_n\in{A}$.

 

[Ackbach]

Well I am trying to show that if $f:{X}\rightarrow{Y}$ and $g:{X}\rightarrow{Y}$ are continuous functions such that $f(x)=g(x)$ for all $x\in{A}$ then $f(x)=g(x)$ for every $x\in{\bar{A}}$.

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 $x_n\in{A}$.

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 $A$?

 

[dray]

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 $A$?

I don't know, since all we are told is that $A$ is a subset of a metric space $X$. Hence my initial question.

 

[Ackbach]

I don't know, since all we are told is that $A$ is a subset of a metric space $X$. Hence my initial question.

Here's the closure definition. With that information, can you answer the question in post # 2?

 

[dray]

Answer to your post would be "yes".

 

[Ackbach]

Answer to your post would be "yes".

Great! So your answer is...

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

 

[dray]

Thanks.

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

Brain fried!

 

[Ackbach]

Thanks.

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

Brain fried!

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

 

[dray]

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

Ahhh....I needed to include \ before the {. I'm a $LaTeX$ novice.

 

[Ackbach]

Ahhh....I needed to include \ before the {. I'm a $LaTeX$ novice.

Understood. Actually, you would write it as $\LaTeX$. The curly brace is the basic way to organize collections of things inside a math environment. Such things have to be escaped (that is, preceded by a backslash) if you want to display the actual character.

Cheerio.