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$(X, d) $ be a metric space and $\emptyset \neq A \subset X$ and $x, y\in X$.

To show: $|d(x, A) - d(y, A)| \le d(x, y)$

My attempt: \begin{align} &d(x, A) = \inf \{ d(x, a) : a\in A \}\\ &d(x, A) \le d(x, a) \le d(x, y) +d(y, a) \forall a\in A. \end{align}

Hence, \begin{align} d(x, A)&\le\inf\{d(x, y)+d(y, a):a\in A\}=\\ &=d(x, y) +\inf \{d(y, a):a\in A\}=\\ &=d(x, y) +d(y, A).\end{align}

It follows $$d(x, A) -d(y, A) \le d(x, y),$$ and analogously $d(y, A) \le d(y, x) +d(x, A) $, that is $$ d(y, A)-d(x, A) \le d(x, y) .$$

Then combining above two inequalities, we get

$|d(x, A) - d(y, A) |\le d(x, y) $

Is my proof correct?

Is there any logical gaps?

Please add your thoughts. Thanks.

ecrin
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SoG
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1 Answers1

1

1.- I would write $\forall a\in A, d(x, A) \le d(x, a) \le d(x, y) +d(y, a) $ instead of $$"d(x, A) \le d(x, a) \le d(x, y) +d(y, a) \forall a\in A"$$

Otherwise,

1.1- at first reading, one wonders what $a$ is;

1.2- In addition, it would allow you to clarify your thinking by writing, for example$$\boxed{\forall a\in A, d(x, A) \le d(x, a) \le d(x, y) +d(y, a) }\color{red}{(*)}\text { since }d(x,A)\text{ is a minorant of}$$ $$\{ d(x, \alpha) : \alpha\in A \}$$

Note that I used $\alpha$ instead of $a$ to avoid notation clash.

2.- By writing $$"\begin{align} d(x, A)&\le\inf\{d(x, y)+d(y, a):a\in A\}=\\ &=d(x, y) +\inf \{d(y, a):a\in A\}=\\ &=d(x, y) +d(y, A).\end{align}"$$

you write too many things in the same time.

And writing less would allow you to be more sure of your proof that you want rigorous.

2.1- First, $d(x,A)$ is a minorant of $\{d(x, y)+d(y, a):a\in A\}$, according to $\color{red}{(*)}$

2.2- Then, as $\inf $ is the greatest of minorants,$$d(x, A)\le\inf\{d(x, y)+d(y, a):a\in A\}$$

2.3- Then $\inf\{d(x, y)+d(y, a):a\in A\}= d(x, y) +\inf \{d(y, a):a\in A\}$, what you seem to find obvious.

2.4- And finally, you are sure that $$d(x, A)\le d(x, y) +d(y, A)$$and $$d(x, A)-d(y,A)\le d(x, y)$$

3.- I don't see anything to report in the rest of your post.

Stéphane Jaouen
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