Htykuut

I am trying to show that $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ is an algebra.

Here, $A^{circ}$ denotes the interior of A.

I am having trouble showing, in particular, that: $A, B in F$ with $0 in (A^c)^circ, 0 in (B^c)^circ$ implies that $A cup B in F$.

I can't figure out how to put things together to show that $0$ is in either $(A cup B)^{circ}$ or $((A cup B)^c)^{circ}$.

A few (potentially) useful facts I know:

1. $((A cup B)^c)^{circ} subseteq (((A cup B)^{circ})^c)^{circ}$


2. $A^{circ} cup B^{circ} subseteq (A cup B)^{circ}$

As $0 in (A^c)^mathrm{o}$ and $0 in (B^c)^mathrm{o}$, then $0 in (A^c)^mathrm{o} cap (B^c)^mathrm{o}$. But $(A^c)^mathrm{o} cap (B^c)^mathrm{o} = ((A^c)cap (B^c))^mathrm{o}=(A cup B)^mathrm{o}$, since the set interior of a finite intersection is the finite intersection of interiors. Hence $0 in (A cup B)^mathrm{o}$.

$A,B in F$ then we have, wlog, 3 possibilities:

1) $0in A^{circ}$ and $0in B^{circ}$

2) $0in (A^c)^{circ}$ and $0in (B^c)^{circ}$

3) $0in A^{circ}$ and $0in (B^c)^{circ}$

For 1) and 3), we can conclude that $0in A^{circ}cup B^{circ} implies 0in (Acup B)^{circ}$ as said your fact 2. Hence $Acup Bin F$

2) We have that there is some open ball $B_0$ centered at zero with radius $varepsilon$ such that $B_0subseteq A^c$ and some $B'_0$ with radius $varepsilon'$such that $B'_0subseteq B^c$. Then we can construct a ball $B$ centered at zero with radius $min{varepsilon , varepsilon'}$ sucht that $Bsubseteq A^ccap B^c$. Hence $0in (A^ccap B^c)^{circ}$, but $A^c cap B^c = (Acup B)^c$.

To show $A,Bin Fimplies Acup Bin F$.

• $0in A^{circ}$ and $0in B^{circ}$ then $0in A^{circ}cup B^{circ}subseteq (Acup B)^{circ}$




• $0in A^{circ} $ and $0in (B^c)^{circ}$ then since $0in A^{circ}subset (Acup B)^{circ} $ since $Asubset Acup B$. The same is true for $0in (A^c)^{circ} $ and $0in B^{circ}$.




• $0in (A^c)^{circ} $ and $0in (B^c)^{circ}$ then $0in (A^c)^{circ} cap (B^c)^{circ}=(A^ccap B^c)^{circ}subset ((Acup B)^c)^{circ}$