It is false. The second set contains one more element than the first: \(\not{0}\). I suspect that you're asking because the empty set is a subset of any other set. But that doesn't mean that it's a member of every set.

Question

asked 2021-02-27

a:\(2\in\{1,2,3\}\)

b:\(\{2\}\in\{1,2,3\}\)

c:\(2\subset\{1,2,3\}\)

d:\(\{2\}\subset\{1,2,3\}\)

e:\(\{2\}\subset\{\{1\},\{2\}\}\)

f:\(\{2\}\in\{\{1\},\{2\}\}\)

asked 2021-02-01

asked 2021-07-14

B) Let \(A=\begin{bmatrix}1 & 0&1 \\0 & 1&0 \end{bmatrix} , B=\begin{bmatrix}1 & 0&0 \\0 & 0&1 \end{bmatrix} \text{ and } C=\begin{bmatrix}1 & 0 \\0 & 1\\0&1 \end{bmatrix}\)

Find \(\displaystyle{\left({B}\cdot{C}\right)}\cdot{A}\)

asked 2021-07-28