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in my topology course we are going through sequences and convergence of sequence and one exercise in my book says the following.

Let $X=(\mathbb{N}\times\mathbb{N})\cup \{(0,0)\}$ with the following topology: (i) $X\setminus\{(0,0)\}$ has the discrete topology; (ii) $U$ is a neighbourhood of $(0,0)$ if $(0,0)\in U$ and the set $\{n\in\mathbb{N}\cup\{0\} \mid (n,m)\not\in U\}$ is finite for almost all $m\in\mathbb{N}\cup\{0\}$. Show that

(a) $X$ is Hausdorff,

(b) there is no sequence in $X\setminus\{(0,0)\}$ convergent to $(0,0)$.

Part (a) is clear to me, but part (b) is a bit obscure. I understand it has to do with the strange way the neighborhood of $(0,0)$ and I tried to argue by contradiction but didn't reach my goal. Can anybody help me?

Michael Hardy
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Michela
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  • What does "almost all" mean here? – Milo Brandt Jul 01 '17 at 21:12
  • @MiloBrandt all but finitely many. My teacher used to joke about: "almost all students in this class will get an A" being a true statement. He used it for convergence: $x_n \to x$ iff every neighbourhood of $x$ contains almost all terms of the sequence. – Henno Brandsma Jul 01 '17 at 22:06

1 Answers1

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Let $S=(p_x)_{x\in \mathbb N}$ be a sequence in $\mathbb N^2.$ Let $T=\{p_x: x\in \mathbb N\}.$

If $T\cap (\mathbb N \times \{m\})$ is finite for every $m\in \mathbb N$ then $\{(0,0)\}\cup (\mathbb N^2$ \ $T)$ is a nbhd of $(0,0)$ that contains no terms of $S$.

If $T\cap (\mathbb N \times \{m_0\})$ is infinite for some $m_0\in \mathbb N,$ then there are infinitely many $x\in \mathbb N$ such that $p_x\in \mathbb N \times \{m_0\},$ and so $U=\{0,0\}\cup (\mathbb N^2$ \ $(\mathbb N\times \{m_0\})$ is a nbhd of $(0,0)$ such that $p_x\not \in U$ for infinitely many $x\in \mathbb N.$

BTW this shows that this topology is not metrizable. Because $\{(0,0)\}$ is not open, but if $p\in Y$ where $Y$ is metrizable and $\{p\}$ is not open, then there exists a sequence $(p_x)_{x\in \mathbb N}$ in $Y$ \ $\{p\}$ such that $\{x\in \mathbb N:p_x\not \in U\}$ is finite for every nbhd $U$ of $p.$

BTW in a metrizable space $Y,$ a sequence $(f(x))_{x\in \mathbb N}$ converges to $p$ iff $\{p\}=\cap_{y\in \mathbb N}\;Cl(\;\{f(x):x\geq y\}\;).$ In other kinds of spaces this may not be true. For the space $X$ in your Q , let $f:\mathbb N\to \mathbb N^2$ be a bijection. Then $\{(0,0)\}=\cap_{y\in \mathbb N}Cl(\{f(x):x\geq y\})$ but $(0,0)$ has a nbhd $U$ such that $\{x\in \mathbb N: f(x)\not \in U\}$ is infinite.

DanielWainfleet
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  • The last statement is not quite right. As an example, for each $n\in\mathbb{N}$ set $a_{n}$ equal to $0$ if $n$ is odd and equal to $n$ if $n$ is even. Then ${0} = \bigcap_{n\in\mathbb{N}}{\rm cl} {a_{m} : m\geq n}$ but the sequence $(a_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}$ does not converge to $0$. – Dean Miller May 21 '25 at 08:01