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I've been studying general topology this term, and there's one basic fact that I don't get. Let $X$ be a topological space, and A some subset.We defined a neighbourhood of $A$ to be all 'points' $x$ such that there's an open set $O$ such that $x$ belongs to $O$ and $O$ is a subset of $A$ (here, no problem). Yet, we defined closure to be all points $x$ such that $X\setminus A$ is not a neighbourhood of $x$.

Here, I have a problem. I would say that not to be a neighbourhood amounts to say that for every open set $U$ (in $X$), either $x$ belongs to $X$, in which case there's some point $y$ that belongs to $X$ and $A$ at the same time (part 1), or $x$ doesn't belong to $U$, in which case $U$ is a subset of $X\setminus A$.(part 2)

But, in other textbooks (like Munkres's General Topology), the closure of $A$ is a set of $x$ such that for every open set $U$ containing $x$, there's a point in intersection between $U$ and $A$.

Here, I'm confused, because it seems to me that these definitions of closure are not equivalent, the second one only having the part 1 of the first definition and not the part 2.

(PS: I've looked at other threads discussing closure, but nobody seems to have the first definition of closure).

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    Start with a point in the first definition of the closure and show it is in the second definition of the closure, and vice versa – 2017-01-11

1 Answers 1

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We denote the two versions of closure in question as:

$$A_1=\{x\in X: X\setminus A \text{ is not a nbh. of }x\}$$ $$A_2=\bigcap\{U\subseteq X: U \text{ closed}, A\subseteq U\}$$

Let $x\in A_2$ then $x\in U$ for each closed that contains $A$. Assume for contradiction that $X\setminus A$ is indeed a nbh. of $x$. So there exists some open set $O$ st. $x\in O\subseteq X\setminus A$. But then $x\not\in X\setminus O$ and $A\subseteq X\setminus O$ and $X\setminus O$ is closed. Contradicting the fact that $x\in A_2$.

For the second implication, we make use of the contrapositive. Therefore, assume $x\not\in A_2$. Then there must be some closed set $C$ such that $x\not\in C$ and $A\subseteq C$. But then $x\in X\setminus C$, which is open and $X\setminus C\subseteq X\setminus A$. Therefore, $X\setminus A$ is a nbh. of $x$.