# Interior and closure: can they be the same?

My question is about the following:

Let $k$ be a natural number ($k\in\{0,1,2,...\}$). Let $\mathcal{B}_k$ be the family of subsets of $\mathbb{R}$ consisting of $\mathbb{R}$ and all the open intervals $(a,b)$ which contain at most $k$ integers (when $k=0$, that means: intervals $(a,b)$ that contain no integers).

Assume that $k=0$ and we look at the topology $\mathcal{T}_0:= \mathcal{T}(\mathcal{B}_0)$ induced by $\mathcal{B}_0$. Compute the interior and the closure of $A=[0,2)$ in $(\mathbb{R}, \mathcal{T}_0)$.

I think that the interior is $int(A)=(0,1)\cup (1,2)$, but then I thought the closure has to be the same as the interior, because the topology we're in is induced by $\mathcal{B}_0$. Am I correct, or just thinking wrong?

Finally, I have to compute the interior and closure of $$D:= \{ (x,y)\in \mathbb{R}^2 : x^2+y^2\leq 1 \} \subset \mathbb{R}^2$$ in $\mathbb{R}^2$ endowed with the product topology $\mathcal{T}_0 \times \mathcal{T}_{eucl}$. I think this has something to do with the fact that for $k\geq 1$, the topology induced by $\mathcal{B}_k$ on $\mathbb{R}$ is the Euclidean topology (I already proved this earlier). But how to compute these?

Thank you!

• Remember that for any set $A$, $A$ is contained in $cl(A)$. So $0$ must be in $cl(A)$. (This is regarding your first question). – Mauro Dec 6 '15 at 20:10
• oh okay, so my interior of $A$ is correct, but $cl(A)$ has to be $[0,1)\cup(1,2)$? – jbuser430 Dec 6 '15 at 20:11
• Thank you very much! Perhaps you can help me with the second part? I find it very difficult... – jbuser430 Dec 6 '15 at 20:19
• @Mauro Why isn't $2$ in the closure? Why isn't $3$ in the closure? – Akiva Weinberger Dec 6 '15 at 22:07
• As for the question in the title, what are $\rm{int}(\Bbb R)$ and $\rm{cl}(\Bbb R)$ in the standard topology? – Akiva Weinberger Dec 6 '15 at 22:17

For the second part, you should think about what a basis of the product topology would be. From the definition of the product topology, a basic open set would be $V=p_1^{-1}(V_1) \cap p_2^{-1}(V_2)$, where $V_1 \in \mathcal{T}_0$ and $V_2 \in \mathcal{T}_{eucl}$, and $p_1, p_2$ are the projections to each coordinate.
Now, any open neighbourhood $V_1 \in \mathcal{T}_0$ of a point $x$ must contain an interval that contains $x$, and doesn't contain any integers. An open neighbourhood $V_2 \in \mathcal{T}_{eucl}$ of a point $y$ must contain (as you know) an open interval around $y$.
From this, we conclude that a neighbourhood $V \in \mathcal{T}_0 \times \mathcal{T}_{eucl}$ of a point $(x,y)$ must contain an open (in the sense of the euclidean topology) rectangle $R$ such that $(x,y) \in R$ and $R$ contains no points of the form $(k,y')$, where $k \in \mathbb{Z}$ (in particular, if a point is of this form, then it doesn't belong to any open set, except for the whole space).
After this, we can see that $\{(x,y) : x^2 + y^2 = 1\} \cap int(D) = \emptyset$ (any rectangle that contains one of these points will not be inside $D$), and also that any point of the form $(0,y)$ will also not be in the interior of $D$ (there are no open sets that aren't the whole space that contain these points). The rest of $D$ will be its interior, as you can verify by taking a small rectangle around any point.
As for the closure of $A$, if you take a point of the form $(k,y)$, then its only neighbourhood is the whole space, so it is trivially an accumulation point of $A$, and belongs to its closure. For all the other points not contained in $A$, you can take a small rectangle that doesn't intersect $A$, so they will not belong to $cl(A)$.