Proof that T is bounded below with ##inf T = 2M##

[member 731016]

Homework Statement

I have found two alternative proofs to the solutions of the problem below so I am wondering whether my proofs are valid.

Relevant Equations

$inf T = 2M$

My first solution is

Let
$S = \{x_1, x_2, x_3, ..., x_n\}$
$T = \{2x_1, 2x_2, 2x_3, ... 2x_n\}$
$T = 2S$

Therefore, $inf T = inf 2S = 2inf S = 2M$

May someone please know whether this counts as a proof?

My second solution is,

$x ≥ M$
$2x ≥ 2M$
$y ≥ 2M$ (Letting y = 2M)

Let $inf T = N$
Therefore by using definition of infimum,
$N ≥ K$ where $K$ is a lower bound
Let $K = \frac{x}{n}$ where $n > 1$
Suppose $N = 2M$
$N ≥ K$
$N ≥ \frac{x}{n}$
$2M ≥ \frac{x}{n}$

However, I am unsure where to go from here.

Any help greatly appreciated - Chiral.

 

[pasmith]

The first proof is invalid, because it is not given that S is finite (or countable).

Your aim is to show that 2M = 2\inf S is the infimum of T = \{ 2x: x \in S\}. Do it directly: is 2M a lower bound for T? If b > 2M, is b a lower bound for T?

 

[docnet]

Your goal is to show that for all $t\in T$, $2M\leq t$.

 

[WWGD]

Homework Statement: I have found two alternative proofs to the solutions of the problem below so I am wondering whether my proofs are valid.
Relevant Equations: $inf T = 2M$

View attachment 341650
View attachment 341651
My first solution is

Let
$S = \{x_1, x_2, x_3, ..., x_n\}$
$T = \{2x_1, 2x_2, 2x_3, ... 2x_n\}$
$T = 2S$

Therefore, $inf T = inf 2S = 2inf S = 2M$

May someone please know whether this counts as a proof?

My second solution is,

$x ≥ M$
$2x ≥ 2M$
$y ≥ 2M$ (Letting y = 2M)

Let $inf T = N$
Therefore by using definition of infimum,
$N ≥ K$ where $K$ is a lower bound
Let $K = \frac{x}{n}$ where $n > 1$
Suppose $N = 2M$
$N ≥ K$
$N ≥ \frac{x}{n}$
$2M ≥ \frac{x}{n}$

However, I am unsure where to go from here.

Any help greatly appreciated - Chiral.

For one, whether $2x >x$ will depend on the sign of $x$. $8>4$, but not so for $-2, -1$.

 

[nuuskur]

The first argument does not count since that's what you have to prove in the first place. Why can you pull the $2$ in front of the infimum?

Either way $2M$ is clearly a lower bound for $T$ and by assumption we can find $x\in S$ satisfying $x<M+\frac{\varepsilon}{2}$, hence $T\ni 2x < 2M+\varepsilon$.