Dense subset

**particlejohn** · December 3rd 2008, 02:12 AM

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?

**Plato** · December 3rd 2008, 08:44 AM

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 $f\left( {x_n } \right) = g\left( {x_n } \right)\,\& \,\left( {f\left( {x_n } \right)} \right) \to f(x)\,\& \,\left( {g\left( {x_n } \right)} \right) \to g(x)$ .
Do you see the contradiction?

**manjohn12** · December 3rd 2008, 10:38 AM

I see the contradiction. Thanks. However, in my opinion, proof by contradiction is a "cop out" way of proving things (even though it is correct). Because you can prove anything from contradictory statements. Was my direct way ok?

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