Why is $\operatorname{Bd}(A) = \{0, 1\}$ not connected and $\operatorname{int}(\Bbb Q) = \varnothing$ connected?

Question

Define $A= [0,1]$ with interior $\operatorname{int}(A) = (0,1)$. I ask the following:

• Why is $\operatorname{Bd}(A) = \{0, 1\}$ is not connected$

• Why is $\operatorname{int}(\Bbb Q) = \varnothing$ is connected?

Answer 1

Because $$\{0,1\} = (U \cap I) \cup (V \cap I)$$ for $$U := (-1/2,1/2) \qquad \text{and} \qquad V := (1/2,3/2)$$ and $I := [0,1]$ is the disjoint union of two open subsets in $\{0,1\}$ and $\varnothing$ is connected by definition.

Answer 2

1. Take any two $\delta$-neighbourhood ($\delta < 0.5$) at each point of ${\rm Bd}(A)$ separating ${\rm Bd}(A)$. Under the subspace topology induced by ${\rm Bd}(A)$, we can easily see that $\{0\} = (-\delta, \delta) \cap {\rm Bd}(A)$ is open in ${\rm Bd}(A)$. Idem for $\{1\}$.
2. Empty set is connected by definition since it can't even contain one nonempty set.

A topological space $X$ is said to be disconnected if it is the union of two disjoint nonempty open sets. Otherwise, X is said to be connected.