0
$\begingroup$
  • In First Countable space: $$F \text{ is closed iff }\,\,\, \forall x_n \subset F \text{ and } x_n \to x, \text{ then } x∈F$$

Proof. Suppose that $\forall x_n \subset F \text{ and } x_n \to x, \text{ then } x∈F.$

We will show that $F$ is closed. That is show that $F=\bar{F}$. Since $F \subset \bar{F}$, I sufficient to show that $\bar{F} \subset F$.

Let $x \in \bar{F}.$ Note that in First countable space has the theorem about, $x \in \bar{F}$ iff there exists $\{x_n\}$ in $F$ such that $x_n \to x.$ By assumption, we have $x∈F.$ Please check the proof right. ??

And (→) I can't prove. Please hint me to proof that.

  • If $f,g : X \to Y$ are continuous and $f|_D = g|_D$ where $\bar{D} = X,$ then $f = g.$

Proof. I can't prove it, please hint me to proof that.

  • 0
    A special case of your second result for metric spaces is shown here: [Continuous functions between metric spaces are equal if they are equal on a dense subset](http://math.stackexchange.com/questions/202325/continuous-functions-between-metric-spaces-are-equal-if-they-are-equal-on-a-dens). The proof given there works for first-countable spaces, too.2012-11-20

3 Answers 3