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  • 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.

Martin Sleziak
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Tee Mth
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3 Answers3

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(i) Let $F$ be closed, and $x_n \in F$ with $x_n \to x$. If you had $x \in X \setminus F$, finally $x_n \not\in F$ (do you know why?).

(ii) This does not hold true without $Y$ being $T_2$. Then you know that $\Delta_Y = \{(y,y) \mid y \in Y\}\subseteq Y^2$ is closed, hence is $(f,g)^{-1}(\Delta_Y)$, where $(f,g)\colon X \to Y^2$ is given by its components $f$ and $g$.

martini
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What you have for the first problem is correct. In the other direction, suppose that $F$ is closed. Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $F$ converging to $x$; we want to show that $x\in F$. Suppose that $x\notin F$. Let $U=X\setminus F$; $F$ is closed, so $U$ is an open neighborhood of $x$. Thus, there is an $m\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m$. Why is this impossible?

The second theorem isn’t always true: you need to assume that $Y$ is Hausdorff. Then suppose that $f\ne g$; that means that there is some point $x\in X$ such that $f(x)\ne g(x)$. $Y$ is Hausdorff, so there are disjoint open sets $U$ and $V$ in $Y$ such that $f(x)\in U$ and $g(x)\in V$. Let $W=f^{-1}[U]\cap g^{-1}[V]$; $W$ is an open neighborhood of $x$ (why?), so there is a point $z\in D\cap W$. Where are $f(z)$ and $g(z)$? Why is this impossible?

Brian M. Scott
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  1. I think, First countability should be used in a more original form, that is, that every point has a countable base of open neighborhoods. What you used is a tautology, since you use the same statement as definition (and, as you wrote it down, the other direction is also trivial). What does $x_n\to x$ mean? That, every open set $U\ni x$ contains almost all $x_n$ (for all $n\ge N_0$, starting from an index $N_0$). From $x_n\in F$, conlude that $x\in \bar F$.
  2. First countablility is used only in the other direction. Let $F$ be a set satisfying the RHS, I would try to prove that $X\setminus F$ is open: take a point $x\in X\setminus F$, and try to show --using the countable base-- that there is an open neighborhood of $x$ disjoint to $F$.
  3. Consider the set $\{x\mid f(x)\ne g(x)\}$ and prove that this is open, and by hypothesis, disjoint to the dense $D$, hence is $\emptyset$.
Berci
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  • You might be mixing up first countability and second countability: First countability says that every point has a countable neighbourhood basis, while second countability says that the space has a countable base of open sets. – T. Eskin Nov 20 '12 at 11:07
  • Ah, yes, thank you. Corrected. – Berci Nov 20 '12 at 11:09