Limit of Subsequence equals Limit of Sequence/Normed Vector Space
Jump to navigation
Jump to search
Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a sequence in $X$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
- $\sequence {x_{n_r} }$ is convergent and $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$
That is, the limit of a convergent sequence equals the limit of a subsequence of it.
Proof
Let $\epsilon > 0$.
Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
- $\exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers‎:
- $\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
- $\norm {x_{n_r} - l} < \epsilon$
The result follows.
$\blacksquare$