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The example is following:

Let $Y=\{(0,y):y \in R\}$. Let $E \subset R^2$, i.e., the subset of the real plane, and $E=Y\cup \{ (\frac 1n, \frac k{n^2}): n\in Z^+, k \in Z\}$. The topology on $E$ is this:

  1. The point $(\frac 1n, \frac k{n^2})$ is open;
  2. $\{U_n(y_0): n=1,2,...\}$ are the nbhds of the point $(0,y_0)$ of $Y$ is defined as following: $$U_n(y_0)=\{(x,y): x \le \frac 1n, |y-y_0|\le x\}$$.

My text book said $Y$ is closed discrete in $E$. I could see $Y$ is closed: for any point $y \in Y^C$, the set $\{y\}$ is open which is disjoint with $Y$. However, I fail to show that $Y$ is a discrete space.

Could anybody help me? Thanks ahead:)

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1 Answer 1

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Given any $y_0 \in \mathbb{R}$ and $n \in \mathbb{Z}^+$, if $(0,y) \in U_n(y_0)$ then we must have $| y - y_0 | \leq 0 \leq \frac{1}{n}$, and therefore $y = y_0$. It follows that $U_n (y_0) \cap Y = \{ (0,y_0) \}$ for all $y_0 \in \mathbb{R}$ and $n \in \mathbb{Z}^+$.

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Mm, It seems so simple. Thanks for your answer. –  Paul Aug 1 '12 at 7:50

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