Let $A = (0,1) \cup [2,3)$ be a subset of $(\mathbb R, \mathfrak T_C)$

$\mathfrak T_C = \{(a,\infty) :a \in \mathbb R\} \cup \{\mathbb R, \emptyset\}$

I need to find the following sets:




Bd(A) I know the basic defintions of all of these sets and I feel like I understand how they are related. However, I get confused with the different topologies and how that affects the final set.

I think the $Int(A) = (0,1) \cup (2,3)$, $Cl(A) = (0,3)$ $Bd(A)=\{0,1,2,3\}$ and the $Ext(A)=(-\infty,0] \cup [1,2] \cup [3, \infty)$

How am I doing?

  • $\begingroup$ Why do I always forget I can use those statements! $\endgroup$ – user219081 May 8 '15 at 19:44

Something is wrong. If your interior and closure are correct, then your boundary is wrong, as $$\begin{align}\text{Bd}(A) = \overline{A} \setminus A^\circ \\ = (0,3)\setminus\left((0,1)\cup(2,3)\right) \\ = [1,2] \\ \neq \{0,1,2,3\}\end{align}$$

Remember that the interior of a set $A$ is the union of all open sets that are subsets of $A$. $A$ is a bounded set while all open sets in $\mathfrak{T}_c$ are unbounded, meaning no open set in the topology is a subset of $A$. Hence, $\text{Int}(A) = \emptyset$.

The closure of $A$ can be obtained from the intersection of all closed sets of which $A$ itself is a subset. A more "hand-wavey" definition is the "smallest" closed set that contains $A$. The closed sets in this topology look like $(-\infty,a]$, so $(-\infty,3]$ should be the closure of $A$. From here you can use the definitions $\text{Bd}(A) = \overline{A}\setminus A^\circ$ and $\text{Ext}(A) = \mathbb{R}\setminus \overline{A}$ to answer the rest of the question.

  • $\begingroup$ so by that definition of boundary, the boundary and the closure are the same set? and the exterior is $(3, \infty)$ $\endgroup$ – user219081 May 8 '15 at 20:01
  • $\begingroup$ Yes to both, assuming my work is correct! $\endgroup$ – graydad May 8 '15 at 20:04

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