Let v be a vector over a field

To evaluate the truth value of the statements, we need to consider the properties and definitions of vectors and subsets in a vector space. Let's examine each statement individually:
Statement 1: If $\mathbf{0}\in S$, then $S$ is a subspace of $V$.
This statement is true. For $S$ to be a subspace of $V$, it must satisfy three conditions:
1. $\mathbf{0}\in S$ (the zero vector is an element of $S$),
2. $S$ is closed under vector addition (if $𝐮$ and $𝐯$ are in $S$, then $𝐮+𝐯$ is also in $S$), and
3. $S$ is closed under scalar multiplication (if $𝐮$ is in $S$ and $c$ is a scalar, then $c𝐮$ is also in $S$).
Statement 2: If $S$ is a subspace of $V$, then $\mathbf{0}\in S$.
This statement is true. As mentioned in the explanation of Statement 1, for $S$ to be a subspace of $V$, it must contain the zero vector $\mathbf{0}$.
Statement 3: If $S\cup T=\{\mathbf{0}\}$, then $S$ and $T$ are subspaces of $V$.
This statement is false. To demonstrate this, we can provide a counterexample. Consider $V$ as the vector space ${ℝ}^2$ and let $S=\{(1,0)\}$ and $T=\{(0,1)\}$. Both $S$ and $T$ are subsets of $V$, and their union is $\{(1,0),(0,1)\}$, which does not equal $\{\mathbf{0}\}$. However, $S$ and $T$ are not subspaces of $V$ because they do not contain the zero vector $\mathbf{0}$.
To summarize:
- Statement 1 is true.
- Statement 2 is true.
- Statement 3 is false.