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This is exercise $1.5$A from Engelking's book, page $47$.

Verify that $X$ is $T_{0}$ if and only if $\overline{\{x\}} \neq \overline{\{y\}}$ for every pair of distinct points $x,y$.

My try:

Suppose first that $X$ is $T_{0}$ and let $x,y$ be two distinct points of $X$. Then since $X$ is $T_{0}$ wlog we can find an open set $U$ such that $x \in U$ and $y \not \in U$. Let us show that $\overline{\{x\}} \neq \overline{\{y\}}$. Suppose otherwise, then $x \in \overline{\{x\}} \subseteq \overline{\{y\}}$. This in turn implies that $U \cap \{y\} \neq \emptyset$ so that $y \in U$, a contradiction.

For the other direction, let $x,y$ be distinct points of $X$. Then either $\overline{\{x\}}$ is not contained in $\overline{\{y\}}$ or $\overline{\{y\}}$ is not contained in $\overline{\{x\}}$. Assume the former, then $x \not \in \overline{\{y\}}$ so we can find an open set $U$ such that $U$ contains $x$ and $U$ does not intersects $\{y\}$. If the other case happens we can find an open set $V$ such that $V$ contains $y$ and $x \not \in V$. Therefore $X$ is $T_{0}$.

Is this OK?

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1. How do you justify ‘this in turn implies ...’? 2. How do justify ‘then $x \notin \overline{\{ y \}}$’? –  Zhen Lin Aug 7 '11 at 10:48
    
@Zhen Lin: $x \in \overline{\{y\}}$ and $U$ is an open set containing $x$ therefore $U$ intersects $\{y\}$. For the other one: note that if $x \in \overline{\{y\}}$ then $\{x\} \subset \overline{\{y\}}$, taking closures on both sides yields $\overline{\{x\}} \subset \overline{\{y\}}$, a contrdiction. –  user10 Aug 7 '11 at 10:52
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@André Nicolas: can you please add your reply as answer so I can accept it? –  user10 Aug 7 '11 at 10:56

1 Answer 1

up vote 3 down vote accepted

The argument is detailed, clear, and correct. Maybe, for added clarity, in the second half of the proof, one could insert "Then $\overline{\{x\}} \ne \overline{\{y\}}."$ between the first two sentences, and begin the next sentence with "So". Or maybe that's overkill.

Added: Some additional explanation of the "This in turn" in the first half of the proof would be helpful. After all, the game here is checking everything, these weak separation properties do not allow normal use of geometric intuition.

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