# Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!

Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$.

I think this statement is false because the definition of closure does have the union of A and the limit points of A. However, I know that it is possible to have boundary points that are not limit points.

If I am correct what is a basic example that models this?

• What is your definition of $\text{Bd}(A)$? Commented Apr 30, 2015 at 14:51
• Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of $A$ if every open set containing $x$ intersects both $A$ and $X−A$.
– user219081
Commented Apr 30, 2015 at 17:23

This is not a false claim. By (one) definition of boundary we have $\text{Bd}(A) = \overline{A}\setminus A^\circ$ where $A^\circ$ is the interior of $A$. Hence $$A \cup (\overline{A}\setminus A^\circ) = \overline{A}$$

We have equality from the union above since $A^\circ \subset A$. Hence $$A\cup \text{Bd}(A) = \text{Cl}(A)$$

Edit: As requested, here is how you could prove this result with sets, elements and double containment.

Let $x \in \overline{A}$. It should be clear that either $x \in A^\circ$ or $x \notin A^\circ$. In the latter case, this means $x\in \overline{A}/A^\circ$ so clearly $x\in A\cup \overline{A}/A^\circ$. In the former case, we can use the fact that $A^\circ \subset A$ to see that $x \in A$ so $x \in A \cup \overline{A}/A^\circ$. Hence, $\overline{A} \subset A\cup \overline{A}/A^\circ$.

Now let $y \in A\cup \overline{A}/A^\circ$ and suppose FTSOC that $y \notin \overline{A}$. Since $A,A^\circ \subset \overline{A}$ this means $y \notin A$ and $y\notin A^\circ$, so $x\notin A \cup \overline{A}$. It is obvious that $A\cup \overline{A} /A^\circ \subset A \cup \overline{A}$ so $y \notin A\cup \overline{A} /A^\circ$, a contradiction. Thus $\overline{A} = A \cup \overline{A}/A^\circ$.

• The boundary is equal to the closure minus the interior, not the closure minus the whole set... Commented Apr 30, 2015 at 14:56
• @NajibIdrissi How does it look now? Commented Apr 30, 2015 at 14:57
• I think he has a different definition of boundary, since he states that there can be points in the boundary that are not limit points.. Commented Apr 30, 2015 at 15:04
• @MohamedHashi the definition may be different, but it must still be equivalent. Commented Apr 30, 2015 at 15:05
• Is he necessarily a 'he'? Commented Apr 30, 2015 at 15:17

You might be getting confused with the following: given a topological space $A$, it is not necessarily the case that every point of the closure $\overline{A}$ is the topological limit of some sequence $a_1,a_2,a_3,\dots$ of points of $a$.

Indeed, let $R$ be an uncountable set, and consider the space $X=\{f\colon R\to\{0,1\}\}$. For a countable set $B\subset A$ and a function $g\colon A\to\{0,1\}$, we define $$U_{B,g}=\{f\in X\;\colon\; f(\alpha)=g(\alpha)\textrm{ for all }\alpha\in B\}$$ The sets $U_{B,g}$ form the basis for a topology on $X$. Now define a subset $Y$ of $X$ by $$Y=\{f\in X\;\colon\;f(\alpha)=0\textrm{ for all but countably many }\alpha\in A\}$$ Now if $g_1,g_2,g_3,\dots$ is a sequence in $Y$ such that $g_n\to g\in X$, then $g\in Y$. So $Y$ contains all its limit points. On the other hand, $Y$ is not closed, since it is dense in $X$.

• I think you meant that set of all limit points of $A$ need not be the closure of $A$. We always have $A\subseteq A^L \subseteq \bar A$, where $A^L$ denotes the set of limit points of convergent sequences in $A$. Commented Apr 30, 2015 at 15:22
• @Ennar Yes I did. Commented Apr 30, 2015 at 15:23