Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:)

share|cite|improve this question
up vote 1 down vote accepted

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 $n \in \mathbb{Z}^+$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.