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Dense Subset in a Metric Space

dray

Member
Feb 5, 2012
37
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 5, 2012
37
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 5, 2012
37
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 5, 2012
37
Answer to your post would be "yes".
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 5, 2012
37
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Member
Feb 5, 2012
37
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Ahhh....I needed to include \ before the {. I'm a $LaTeX$ novice.
Understood. Actually, you would write it as $\LaTeX$. (Nod) 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.