MHB Necessary and Sufficient Meaning of Isometries by D. J. H. Garling

[Math Amateur]

Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help in order to understand some remarks by Garling made at the start of Section 11.5 ... ...

The remarks by Garling made at the start of Section 11.5 ... ... read as follows:
View attachment 8977In the above remarks Garling talks abut "the condition" being necessary and "the condition" being sufficient ...

It seems to me that that "the condition" is as follows:

$$T$$ is an isometry $$\Longleftrightarrow \| T(x) \|_F = \| x \|_E \text{ for all } x \in E$$Can someone explain to me the meaning of Garling's use of necessary and sufficient ...

NOTE: It seems that in the case where Garling says the condition is necessary that he is proving ...

$$T$$ is an isometry $$\Longrightarrow \| T(x) \|_F = \| x \|_E \text{ for all } x \in E $$

and when Garling says he is proving sufficiency he is proving ...

$$ \| T(x) \|_F = \| x \|_E \text{ for all } x \in E \Longrightarrow T$$ is an isometry ...

But why is this ... I need to fully understand necessity and sufficiency ... Hope someone can help ...

Peter***NOTE 2***From what I understand in basic logic ...

$$P \Longrightarrow Q$$ in words means $$P$$ is sufficient for $$Q$$ ...

while $$\sim P \Longrightarrow \sim Q$$ translates to $$P$$ is necessary for $$Q$$ ...

 

[Opalg]

Re: Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

Can someone explain to me the meaning of Garling's use of necessary and sufficient ...

NOTE: It seems that in the case where Garling says the condition is necessary that he is proving ...

$$T$$ is an isometry $$\Longrightarrow \| T(x) \|_F = \| x \|_E \text{ for all } x \in E $$

and when Garling says he is proving sufficiency he is proving ...

$$ \| T(x) \|_F = \| x \|_E \text{ for all } x \in E \Longrightarrow T$$ is an isometry ...

But why is this ... I need to fully understand necessity and sufficiency ...
From what I understand in basic logic ...

$$P \Longrightarrow Q$$ in words means $$P$$ is sufficient for $$Q$$ ...

while $$\sim P \Longrightarrow \sim Q$$ translates to $$P$$ is necessary for $$Q$$ ...

As you correctly say, $$\neg P \Longrightarrow \neg Q$$ translates to $$P$$ is necessary for $$Q$$. But $$\neg P \Longrightarrow \neg Q$$ is equivalent to $Q \Longrightarrow P$. So $$Q \Longrightarrow P$$ in words means $$P$$ is necessary for $$Q$$.

 

[Evgeny.Makarov]

Re: Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

It seems to me that that "the condition" is as follows:

$$T$$ is an isometry $$\Longleftrightarrow \| T(x) \|_F = \| x \|_E \text{ for all } x \in E$$

The condition is $$\| T(x) \|_F = \| x \|_E$$ for all $$x \in E$$. The fact that it is necessary for $T$ to be an isometry means that if $T$ is an isometry, then the equality holds. The fact that it is sufficient means that it implies that $T$ is an isometry.

 

[Math Amateur]

Re: Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

The condition is $$\| T(x) \|_F = \| x \|_E$$ for all $$x \in E$$. The fact that it is necessary for $T$ to be an isometry means that if $T$ is an isometry, then the equality holds. The fact that it is sufficient means that it implies that $T$ is an isometry.

Thanks to Opalg and Evgeny I think I now am clear on the issues above...To summarise ...

The condition, say $$P \equiv \| T(x) \|_F = \| x \|_E \text{ for all } x \in E$$ ...

Let $$Q \equiv$$ $$T$$ is an isometry ... ...
Then ... $$P$$ is necessary for $$Q$$ ... means ...

$$\sim P \Longrightarrow \sim Q$$ ... which is equivalent to ...

$$Q \Longrightarrow P$$ ... which means ... $$T$$ is an isometry $$\Longrightarrow \| T(x) \|_F = \| x \|_E \text{ for all } x \in E$$ ...

... and ...

$$P$$ is sufficient for $$Q$$ ... means ...

$$P \Longrightarrow Q$$ ... which means ...

$$\| T(x) \|_F = \| x \|_E \text{ for all } x \in E \Longrightarrow$$ $$T$$ is an isometry ...
Is the above correct ...Thanks again for your help ...

Peter

 

[Evgeny.Makarov]

Re: Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

Yes, it is correct.

 

[Math Amateur]

Re: Isometries ... Remarks by Garling including terms "necessary" and "sufficient" ...

Yes, it is correct.

Thanks Evgeny ...

Appreciate your help ...

Peter