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Prove that if $(X,\rho)$ is a metric space, then the function $d(x,y)=\rho(x,y)+1$ is also a metric on $X$ . In other words, the question is asking you to prove that if you take a metric space and add $1$ to the distance between any two points in that space, the resulting function will also be a metric space.

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Write and solve direct variation equations

Full solution

Q. Prove that if

(X,\rho)

is a metric space, then the function

d(x,y)=\rho(x,y)+1

is also a metric on

X

. In other words, the question is asking you to prove that if you take a metric space and add

1

to the distance between any two points in that space, the resulting function will also be a metric space.

Non-negativity Property: To prove that $d(x,y) = \rho(x,y) + 1$ is a metric on $X$ , we need to show that it satisfies the four properties of a metric: $\newline$ $1$ . $d(x,y) \geq 0$ for all $x, y$ in $X$ (non-negativity) $\newline$ $2$ . $d(x,y) = 0$ if and only if $x = y$ (identity of indiscernibles) $\newline$ $3$ . $d(x,y) = d(y,x)$ for all $x, y$ in $X$ (symmetry) $\newline$ $4$ . $X$ $0$ for all $X$ $1$ in $X$ (triangle inequality)
Identity of Indiscernibles Property: First, we check the non-negativity property. Since $\rho$ is a metric on $X$ , we know that $\rho(x,y) \geq 0$ for all $x, y$ in $X$ . Therefore, $d(x,y) = \rho(x,y) + 1$ will also be greater than or equal to $0$ for all $x, y$ in $X$ , because adding $1$ to a non-negative number keeps it non-negative.
Symmetry Property: Next, we check the identity of indiscernibles property. For any $x$ in $X$ , we have $\rho(x,x) = 0$ because $\rho$ is a metric. Thus, $d(x,x) = \rho(x,x) + 1 = 0 + 1 = 1$ . This shows that $d(x,x) \neq 0$ , which is not exactly what we want. However, if $x \neq y$ , then $\rho(x,y) > 0$ , and so $d(x,y) = \rho(x,y) + 1 > 1$ . This means that $d(x,y) = 0$ if and only if $X$ $0$ is not satisfied because $X$ $1$ is never $X$ $2$ . We have found a math error.

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