SOLUTION: Let W be the set of all sequences which converge to 0. Is W a subspace of the vector space of convergent sequences? You must justify your answer.

Question 1166341: Let W be the set of all sequences which converge to 0. Is W a subspace of the vector space of convergent sequences? You must justify your answer.
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
Yes, **W**, the set of all sequences which converge to $0$, is a **subspace** of the vector space of convergent sequences, which we can call $V$.
The vector space $V$ is the set of all sequences of real numbers, $\{a_n\}_{n=1}^\infty$, such that $\lim_{n \to \infty} a_n = L$, where $L$ is a finite real number.
The subset $W$ is defined as:
$$W = \left\{ \{a_n\} \in V \ \bigg| \ \lim_{n \to \infty} a_n = 0 \right\}$$
To prove that $W$ is a subspace of $V$, we must verify the three subspace conditions:
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## 1. The Zero Vector Condition (Contains the Zero Sequence)
The zero vector in $V$ is the **zero sequence**, $\mathbf{z} = \{z_n\}$, where $z_n = 0$ for all $n \ge 1$.
We check if the zero sequence satisfies the condition for $W$:
$$\lim_{n \to \infty} z_n = \lim_{n \to \infty} 0 = 0$$
Since the limit of the zero sequence is $0$, the zero sequence is in $W$.
Thus, $W$ is non-empty.
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## 2. Closure under Vector Addition ➕
Let $\mathbf{a} = \{a_n\}$ and $\mathbf{b} = \{b_n\}$ be two arbitrary sequences in $W$. This means:
* $\lim_{n \to \infty} a_n = 0$
* $\lim_{n \to \infty} b_n = 0$
We need to check if their sum, $\mathbf{a} + \mathbf{b} = \{a_n + b_n\}$, is also in $W$. We use the **Limit Sum Law**:
$$\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n$$
Substitute the conditions for $\mathbf{a}$ and $\mathbf{b}$:
$$\lim_{n \to \infty} (a_n + b_n) = 0 + 0 = 0$$
Since the sum of the sequences converges to $0$, the sequence $\mathbf{a} + \mathbf{b}$ is in $W$.
Thus, $W$ is closed under sequence addition.
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## 3. Closure under Scalar Multiplication ✖️
Let $\mathbf{a} = \{a_n\}$ be an arbitrary sequence in $W$ (so $\lim_{n \to \infty} a_n = 0$), and let $c$ be an arbitrary scalar (a real number).
We need to check if the scalar multiple, $c\mathbf{a} = \{c \cdot a_n\}$, is also in $W$. We use the **Limit Constant Multiple Law**:
$$\lim_{n \to \infty} (c \cdot a_n) = c \cdot \lim_{n \to \infty} a_n$$
Substitute the condition for $\mathbf{a}$:
$$\lim_{n \to \infty} (c \cdot a_n) = c \cdot (0) = 0$$
Since the scalar multiple of the sequence converges to $0$, the sequence $c\mathbf{a}$ is in $W$.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space of convergent sequences, $V$.
Would you like to analyze a different set of sequences, such as the set of sequences that converge to $1$, to see if it is a subspace?
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