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In Munkres's Topology (section 17 Closed Sets and Limit Points, 2nd edition), Example 7 consider the closure of a subset of a subspace of a topological space.

Example 7. Consider the subspace $Y = (0,1]$ of the real line $\mathbb{R}$. The set $A = (0,\frac{1}{2})$ is a subset of $Y$; its closure in $\mathbb{R}$ is the set $\bar{A} = [0, \frac{1}{2}]$, and its closure in $Y$ is the set $[0, \frac{1}{2}] \cap Y = (0, \frac{1}{2}]$.

I can follow the example in this presentation, that is to say, by Theorem 17.4, which shows that the closure of $A$ in $Y$ equals $\bar{A} \cap Y$.

However, when I check the closure set $(0, \frac{1}{2}]$ against the Theorem 17.5, which gives a sufficient and necessary condition of closure, I am confused with the point $0 \in \mathbb{R}$. My argument is as follows:

My argument of $0 \in \bar{A}:$ First, the open set $U$ in the subspace $Y$ is one of the five forms: $\emptyset$, $(0,b)\; 0 < b \le 1$, $Y = (0, 1]$, $(a,b) \; 0 < a < b \le 1$, and $(a, 1] \; 0 < a < 1$. There is no open set of $Y$ containing the point $0$. Therefore, $0 \in \bar{A}$ because the sufficient condition "every open set $U$ (of $Y$) containing $0$ intersects $A$" are satisfied vacuously.

I know the argument must be wrong, because $0 \notin Y$! However, what is wrong with my argument?

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$Y$ is the universe. $0$ does not exist (for the purposes of closure etc. in $Y$). – Daniel Fischer Jan 17 '14 at 12:16
up vote 3 down vote accepted

$0$ does belong to $\overline{A}$ where this denotes the closure of $A$ in $\mathbf{R}$. However, if by $\overline{A}$ we mean the closure of $A$ in $Y$, then $0 \not\in \overline{A}$. The theorem you are referring to assumes that the point you are checking belongs to the topological space under consideration, even though this is not mentioned explicitly.

Let me add that you're making a mistake in saying that all open sets of $Y$ are of the indicated forms. In fact, those sets form a basis of open sets, but do not constitute all open subsets of $Y$.

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Thx, I see now. I am also appreciated for you pointing out the basis vs. all mistake. – hengxin Jan 17 '14 at 12:26

Following the above, I would like to note that you yourself have given that the set $(\frac{1}{2},1]$ is open in $Y$. It follows that its complement is closed by the definition of a closed subset of a topological space, and so the set $(0,\frac{1}{2}]$ is closed. The closure of a closed subset of a topological space is itself and so $\overline{(0,\frac{1}{2}]}=(0,\frac{1}{2}]$ in $Y$.

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Following your approach, I actually have another problem about closure and closed set. In the proof of Corollary 17.7 (also in Munkres's Topology), it says that The set $A$ is closed if and only if $\bar{A} = A$. I find no where its justification is given. And I can only prove the only if part ($A$ is closed $\Rightarrow$ $\bar{A} = A$, as used in your approach). Could you please give me some hint to prove the opposite direction (i.e., $\bar{A} = A$ $\Rightarrow$ $A$ is closed)? – hengxin Jan 17 '14 at 13:53
Depending on your definition of closure, proving the contrapositive would probably be easiest. Show that if $A$ is not closed, then $\overline{A}$ contains, as a proper subset, $A$. – Dan Rust Jan 17 '14 at 13:57
I don't understand the claim "depending on your definition of closure". What if the closure of $A$ is defined as the intersection of all closed sets containing $A$ (as Munkres did in Topology)? To prove its contrapositive, I failed to take any step from that "if $A$ is not closed". I only have at hand the definition of closed sets as the complements of open sets. What can we say more about a closed set (and not a closed set)? – hengxin Jan 17 '14 at 14:09
If $A\subset X$ is not closed, then let $\{C_\lambda\subset X\mid A\subset C_\lambda,\: C_\lambda\mbox{ closed }\}$ be the set of all closed subset of the space $X$ which contain $A$. By definition, $\bigcap C_\lambda=\overline{A}$. Further, because each $C_\lambda$ is closed, their intersection is closed. As $A$ is not closed, it can not be equal to the closed set $\bigcap C_\lambda=\overline{A}$ and so $A\neq \overline{A}$. – Dan Rust Jan 17 '14 at 14:18
Thanks, I see now. I am a beginner of general topology and I find that topology is so abstract. I have to practice on the basic definitions and theorems over and over. Thanks for your patience. – hengxin Jan 17 '14 at 14:27

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