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?
Best Answer
$0$ 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 $Y$.