Let (M,d) be a metric space, and M1, M2 be two subspace of M.

Edit: I forgot to mension that both M1 and M2 are dense.

Suppose we have

$f: M rightarrow M$

and we know f is an isometric mapping from M1 to M1, i.e.,

$$

d(f(a),f(b))=d(a,b)

$$

for any $a,b in M1$.

Suppose we also have

$$

d(f(a),f(b))=d(a,b)

$$

for any $a in M1$ and $b in M2$.

The question is : can we say that f is an isometric mapping of $M1cup M2$?

In other words, is it possible to show that

$$

d(f(a),f(b))=d(a,b)

$$

for any $a,b in M1cup M2$.

I think it is true in Euclidean space, but I have no idea how to do it in a general metric space.

Any suggestion would be greatly appreciated.

Mathematics Asked by user856180 on December 28, 2020

1 Answers$M_1={a}$, $M_2={b,c}$, $f(a)=a, f(b)=f(c)=b$, $d(x,y)=1$ if $xneq y.$

Answered by Tsemo Aristide on December 28, 2020

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