interior of $]-5, 0]\cup\left\{ \frac{1}{n}: n \geq1 \right\} \cup \{2,4\} $

My question about the interior of $A=]-5, 0]\cup\left\{ \frac{1}{n}: n \geq1 \right\} \cup \{2,4\} $

Let $A_1=]-5, 0]$, $A_2=\left\{ \frac{1}{n}: n\geq1 \right\}$ and $A_3=\{2,4\}$

$\mathring{A_1}=]-5, 0[$.

$A_2=\left\{ \frac{1}{n}: n\geq1 \right\}$ contains no open sets, and thus the interior $\mathring{A_2}=\emptyset$.

$A_3= \{2,4\} $ contains no open sets, and thus the interior $\mathring{A_3}=\emptyset$.

So we have $\mathring{A_1}\cup \mathring{A_2}\cup \mathring{A_3}=]-5, 0[$

Since $A=]-5, 0]\cup\left\{ \frac{1}{n}: n \geq1 \right\} \cup \{2,4\}=A_1\cup A_2\cup A_3$

And $\mathring{A}=\mathring{\overbrace{A_1\cup A_2\cup A_3}}$

But we don't have $\mathring{A_1}\cup \mathring{A_2}\cup \mathring{A_3}=\mathring{\overbrace{A_1\cup A_2\cup A_3}}$

How can I find $\mathring{A}$?

Any help will be appreciated.

Solution 1:

$\mathring{A}\cup \mathring{B}=\mathring{\overbrace{A \cup B}}$

In general the above result is not true.

For example.

$\mathring{\mathbb{Q}}\cup \mathring{({\mathbb{R}\setminus \mathbb{Q}})}\neq \mathring{\mathbb{R} }$

But in your question $\delta(A_1) \cap \delta {(A_2})\cap {\delta(A_3)}=\emptyset$

Where, $\delta({A}) =bdd(A) $

And rest of your proof is ok.

Solution 2:

Just note that $(-5,0)$ is open and the other points of $A$, $0$,$-5$, the $\frac1n$ and $0$ and $2$ all are non-interior to $A$ because all their resp neighbourhoods contain points outside of $A$.

Those observations are enough to conclude that the interior of $A$ is $(-5,0)$. The division in parts is not needed at all.