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math question

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This forum is in serious need of a LaTeX interpreter, but here it goes.

Update: here's an image of the LaTeX output
closedsets.png


LaTeX Source:
To really help you, I need to know what definition of closed sets and boundary you're working with, but here is how I would do it. I'm going to use PlanetMath.org definitions. Also, I'm going to suppose $E$ is a subset of some metric space $M$ with metric $d$.

$$\partial E = cl(E) \cap cl(E^c)$$

Thus, for all $x \in \partial E$, we know for all $\varepsilon > 0$, $D(x,\varepsilon) \cap E \neq \emptyset$ and $D(x,\varepsilon) \cap E^c \neq \emptyset$, where $D(x,r) = \{y \in M \mid d(y,x) < r\}$.

To prove that $E$ is closed, we prove that it's complement $E^c$ is an open set. To do that, we show that for every $x \in E^c$, there exists a radius $r \in \mathbb{R^+}$ such that $D(x,r) \subseteq E^c$.

Well suppose for a contradiction that this is not true. Then, for some $x_0 \in E^c$, we have $D(x_0,r) \not\subseteq E^c$ for all $r \in \mathbb{R^+}$. It follows that $D(x_0,r) \cap E \neq \emptyset$ for all $r$. Thus, $x_0$ is a limit/accumulation point of $E$, so $x_0 \in cl(E)$ by definition of a closure. Moreover $x_0$ is clearly in $cl(E^c)$ since $E^c \subseteq cl(E^c)$ again by definition of a closure. Thus, by definition $x_0 \in \partial E$. This implies $x_0 \in E$ since by assumption $\partial E \subset E$. This is impossible. $x_0$ cannot be both in $E$ and its complement $E^c$. Thus, $E^c$ must be open, so $E$ is closed. \hfill $\blacksquare$
 
Last edited:
blinky8225 said:
This forum is in serious need of a LaTeX interpreter, but here it goes.

Update: here's an image of the LaTeX output
closedsets.png


To really help you, I need to know what definition of closed sets and boundary you're working with, but here is how I would do it. I'm going to use PlanetMath.org definitions. Also, I'm going to suppose $E$ is a subset of some metric space $M$ with metric $d$.

$$\partial E = cl(E) \cap cl(E^c)$$

Thus, for all $x \in \partial E$, we know for all $\varepsilon > 0$, $D(x,\varepsilon) \cap E \neq \emptyset$ and $D(x,\varepsilon) \cap E^c \neq \emptyset$, where $D(x,r) = \{y \in M \mid d(y,x) < r\}$.

To prove that $E$ is closed, we prove that it's complement $E^c$ is an open set. To do that, we show that for every $x \in E^c$, there exists a radius $r \in \mathbb{R^+}$ such that $D(x,r) \subseteq E^c$.

Well suppose for a contradiction that this is not true. Then, for some $x_0 \in E^c$, we have $D(x_0,r) \not\subseteq E^c$ for all $r \in \mathbb{R^+}$. It follows that $D(x_0,r) \cap E \neq \emptyset$ for all $r$. Thus, $x_0$ is a limit/accumulation point of $E$, so $x_0 \in cl(E)$ by definition of a closure. Moreover $x_0$ is clearly in $cl(E^c)$ since $E^c \subseteq cl(E^c)$ again by definition of a closure. Thus, by definition $x_0 \in \partial E$. This implies $x_0 \in E$ since by assumption $\partial E \subset E$. This is impossible. $x_0$ cannot be both in $E$ and its complement $E^c$. Thus, $E^c$ must be open, so $E$ is closed. \hfill $\blacksquare$
Click to expand...

Very nice. We do need a LaTeX parser for the forums.
 
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