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Q1: Suppose $(x_n)$ satisfies $-1$ $\leq$ $x_n$ $\leq$ $1$ for all $n$ $\in$ $\Bbb{N}$. Suppose $(x_{n_k})$ is a subsequence of $(x_n)$ converging to $(x_0$) $\in$ $\Bbb{R}$. Prove that it is always true that $-1$ $\leq$ $x_0$ $\leq$ $1$

Just for confirmation, is this true because we know (x_n) is bounded, thus by Bolzano-Weierstrass Theorem (x_n) has a monotone subsequence $(x_{n_k})$ and this monotone sequence is either non-increasing or non-decreasing. If it is non-increasing then the sequence $x_0$ is equal to the infimum of the set, and if it is non-decreasing then $x_0$ is equal to the supremum of the set?

Q2: Suppose $(x_n)$ satisfies $-1$ $\leq$ $x_n$ $\lt$ $1$ for all $n$ $\in$ $\Bbb{N}$. Suppose $(x_{n_k})$ is a subsequence of $(x_n)$ converging to $(x_0$) $\in$ $\Bbb{R}$. Prove that it is always true that $-1$ $\leq$ $x_0$ $\leq$ $1$.

How can this be the case?

Q3 Is the statement $\alpha$ $\in$ $\Bbb{R}$ and x $\in$ $\Bbb{R^d}$, then $\Vert{\alpha x}$$\Vert$ $=$ $\alpha$ $\Vert{x}$$\Vert$ true or false?:

Why must the RHS of the equation have $\vert{\alpha}$$\vert$ and not just $\alpha$

Q4: Let $x$, $y$ $\in$ $\Bbb{R^d}$ Prove using the triangle inequality $\Vert{x-y}$$\Vert$ + $\Vert{x - 2y}$$\Vert$ $\geq$ $\Vert{y}$$\Vert$.

Not sure how to prove this?

Sam Connell
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1 Answers1

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Q1 and Q2 appear to be the same. They are a consequence of the following fact:

Lemma. If $x_n \leq M, \forall n \in \mathbb{N}$ for some real $M$ and $(x_n)$ converges to a limit $l$, then $l \leq M$.

Proof. Suppose for a contradiction that $l > M$. Then $\epsilon = \frac{l-M}{2}>0$. Since $(x_n)$ converges to $l$, there is an $N > 0$ such that $$\mid x_n-l\mid < \epsilon, \forall n > N.$$

This implies that $x_n > l - \epsilon = \frac{l+M}{2} > M$, which is false. Hence, we need $l \leq M$.

We get a similar result for a lower bound.

Q3. The LHS is always positive so we need an absolute value on the RHS to have it always be positive.

Q4. $\Vert y \Vert = \Vert (x-y) - (x-2y) \Vert$. Now apply Triangle Inequality.

Anonymath
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