# Dense subset

• Dec 3rd 2008, 02:12 AM
particlejohn
Dense subset
Let $X$ and $Y$ be metric spaces and let $D$ be a dense subset of $X$. Let $f: X \to Y$ and $g: X \to Y$ be continuous functions. Suppose that $f(r) = g(r)$ for all $r \in D$. Prove that $f = g$.

Proof. Let $r \in X \backslash D$. Then $r$ is a limit point of $D$ by the denseness of $D$. Pick a decreasing sequence of positive real numbers $(x_n)$. Then in each open ball $B_{x_{n}}(r)$, there is a point $q_n \neq r$ such that $q_n \in D$. Form these terms as a sequence $(q_n)$. Fix $\epsilon > 0$. Since $x_n \to 0$ any open ball centered at $0$ contains at least one $x_j$ for some $j \in \mathbb{N}$. So there is some $j$ such that $x_j \in B_{\epsilon}(0)$. Then for $n \geq j$, $q_{n} \in B_{\epsilon}(r)$. So $q_n \to r$. Then $g(r) = \lim_{n \to \infty} g(q_n)$. We know that $q_n \in D$ for $n \in \mathbb{N}$. Thus $f(q_n) = g(q_n)$ which implies $f(r) = g(r)$. Thus $f = g$. $\blacksquare$

Is this correct?
• Dec 3rd 2008, 08:44 AM
Plato
Quote:

Originally Posted by particlejohn
Let $X$ and $Y$ be metric spaces and let $D$ be a dense subset of $X$. Let $f: X \to Y$ and $g: X \to Y$ be continuous functions. Suppose that $f(r) = g(r)$ for all $r \in D$. Prove that $f = g$.

First, you must do away with thinking of a general metric space as the real numbers alone. That is, a decreasing sequence of points has no general meaning.

Now suppose that $\left( {\exists x} \right)\left[ {f(x) \ne g(x)} \right]$ clearly $x \in X\backslash D$.
As you noted, by density there is a sequence of distinct points $\left( {x_n } \right) \to x\,\&\,x_n \in D$.
$\left( {\exists \delta > 0} \right)\left[ {B_\delta \left( {f(x)} \right) \cap B_\delta \left( {g(x)} \right) = \emptyset } \right]$.
But $f\left( {x_n } \right) = g\left( {x_n } \right)\,\& \,\left( {f\left( {x_n } \right)} \right) \to f(x)\,\& \,\left( {g\left( {x_n } \right)} \right) \to g(x)$.