# A finite subset of a metric space

• Jan 10th 2009, 04:56 PM
aliceinwonderland
A finite subset of a metric space
Show that a finite subset of a metric space has no limit points and is therefore a closed set.
• Jan 10th 2009, 06:44 PM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
Show that a finite subset of a metric space has no limit points and is therefore a closed set.

A subset of a metric space is closed if and only if it contains its limit points.
To be a limit point any neighborhood needs to contains infinitely many points within the set.
Thus, a finite subset cannot have any limit points.
Thus, the set of limit points is the empty set.
But, the finite subset contains the empty set.
Thus, the finite subset is closed.
• Jan 10th 2009, 06:48 PM
Mathstud28
Quote:

Originally Posted by aliceinwonderland
Show that a finite subset of a metric space has no limit points and is therefore a closed set.

This is an alternate solution to TPHs...it is a little more formal. Note, you may want to wait for confirmation of my solution by a more knowledgable member (Nod)

Let $X$ be a metric space and let $\left\{p\right\}=A\subset X$. Now the only possibilities for a limit point of $A$ is the point $p$, but note that since there is only one point $A$ no matter how small you make an open ball around $p$ there will not be another point of $A$, thus a singleton in $X$ has no limit points, thus it is closed and consequently its complement is open. So now let another set $\left\{x_1,x_2,\cdots,x_n\right\}=B\subset X$ where $n$ is finite. It is clear that $B=\bigcup_{i=1}^{n} x_i$ we will now use the following lemma

Lemma: The intersection of any finite number of open sets is itself open.

Proof: Let $U_i~~1\leqslant i\leqslant n$ be open, then let $x\in\bigcap_{i=1}^{n}U_i$. This implies that $x\in U_i~\forall i$, and since each $U_i$ is open there exists an open ball $\mathcal{O}_{\varepsilon_i}(p,X)$ corresponding to each $sU_i$ such that $\mathcal{O}_{\varepsilon_i}(p,X)\subset U_i$. So now let $\varepsilon=\min\left\{\varepsilon_1,\varepsilon_2 ,\cdots\right\}$, then $\mathcal{O}_{\varepsilon}(x,X)\subset\bigcap_{i=1} ^{n}U_i\quad\blacksquare$

Using the above lemma and the fact that the complement of any singleton in $X$ is open we can see that the complement of any finite point set $\left[\bigcup_{i=1}^{n}x_i\right]^c=\bigcap_{i=1}^{n}x_i^c$ is open, thus the finite set is closed.