$T_n$ it is equal to the closure of its interior: $T_n=\bar{\stackrel{\circ}{T_n}}$

For any subset $Y$ of a topological space $T$ , the biggest open set in $Y$ is called the interior of $Y$ [4][5], and is typed by $\stackrel{\circ}{Y}$ . In others words,

$$\stackrel{\circ}{Y}=\bigcup_{j\in J}X_j$$ (5)

Where $\{X_j;j\in J\}$ is the set of open subsets in $Y$ . A graphical pattern $T_n\in\stackrel{\circ}{Y}$ is included in its interior only when there exists some $X_j$ subset such that $X_j\subset Y$ and $T_n\in X_j$ .

For any subset $Y$ in a topological space $T$ , the smallest closed set that contains the subset $Y$ , is called the closure of $Y$ [4][5], and we will call it $\bar{Y}$ . In others words,

$$\bar{Y}=\bigcap_{j\in J}F_j$$ (6)

Where $\{F_j;j\in J\}$ is the set of all closed subsets which contains the subset $Y$ . In a similar way, a graphical pattern $T_n$ is included in the closure of $Y$ only when all subsets which contain the subset $Y$ has to $T_n$ as one of its elements.

Now, considering the unitary set $T_n=\{t_n\}$ , let us see which is its interior. Let be $\stackrel{\circ}{T_n}$ the interior of $T_n$ . The biggest open set in $t_n$ is $t_n$ too, because is the only element in $T_n$ . Then $\stackrel{\circ}{t_n}=t_n$ . Next the closure of $T_n$ is the biggest open set contained in $t_n$ is $t_n$ , let us see if it is closed.

Taking again the topological descriptions used up to now, $(T,\mathbb{P}(T))$ is the topological space, and we can note that for any $C\in\mathbb{P}(T)$ set, $C-T_n \in \mathbb{P}(T)$ . That is because all the subsets of $T$ are contained in $\mathbb{P}(T)$ , so, $T_n$ is closed; and we have again $\bar{T_n}=T_n$ . Then $\bar{\stackrel{\circ}{T_n}}=T_n$ . This extra condition is necessary to make sure that there exist at least one way to cover the 2-space, occupied by a triangle $T_n$ , for a fixed $n$ .