let $f : ( X , d_1) \rightarrow (Y, d_2)$ be a homeomorphism between two metric spaces. Then choose the correct option
$1. f(A)$ is bounded subset of $Y$ , whenever $A$ is a bounded subset of $X$.
$2.$ $f (A^{\circ}) = (f (A))^{\circ}$ , for $A \subseteq X$, where $A^{\circ}$ denote the interior of $A$
$3.$ $f(\bar A) = \overline{f(A)}$ for $A \subseteq X$, where $\bar A$denotes the closure of $A$
$4$.$d_1(x,y) = d_2(f(x), f(y))$ for all $x, y \in X$
My attempt : I take $ (X, d_1)= (X, d_2) = (\mathbb{R} , d)$ where $d= |x-y|$
Then i thinks option $1$ is false take $f(x) = \frac{1}{1+x^2}$ but A = $\mathbb{R}$ which unbounded set
For option $2$ is true ,take $f(x) = x$ where $x= \mathbb{Q}$ we know that $\mathbb{Q}^{\circ}= \phi$
option $3$ is false take $f(n) = \frac{1}{n}$ where $A= \mathbb{N}$
option $4$ is true by definition of metric space
Is its true ?
Best Answer
Homeomorphism between two metric spaces simply means their topology is same. Options 1 and 4 are false as they are based on metric properties. Option 4 is the condition for isometry. Under homeomorphism, the properties will be preserved which are described in terms of open sets and not in terms of distance.