Question:

Step 1: Understand the given relation The relation $R$ is defined on $\mathbb{N}\times\mathbb{N}$ such that $(x_1,y_1)\,R\,(x_2,y_2)$ if and only if $x_1\leq x_2$ or $y_1\leq y_2$. We need to check for reflexivity, symmetry, and transitivity. Step 2: Check for Reflexivity A relation $R$ is reflexive if for every element $a$ in its domain, $(a,a) \in R$. For this relation, an element is an ordered pair $(x,y) \in \mathbb{N}\times\mathbb{N}$. We need to check if for any $(x,y) \in \mathbb{N}\times\mathbb{N}$, $(x,y)\,R\,(x,y)$ holds. According to the definition, $(x,y)\,R\,(x,y)$ if $x \leq x$ or $y \leq y$. Since $x \leq x$ is always true for any natural number $x$, the condition "$x \leq x$ or $y \leq y$" is always satisfied. Thus, $R$ is reflexive. Step 3: Check for Symmetry A relation $R$ is symmetric if for every $a, b$ in its domain, if $(a,b) \in R$ then $(b,a) \in R$. Here, we need to check if for any $(x_1,y_1), (x_2,y_2) \in \mathbb{N}\times\mathbb{N}$, if $(x_1,y_1)\,R\,(x_2,y_2)$, then $(x_2,y_2)\,R\,(x_1,y_1)$. Assume $(x_1,y_1)\,R\,(x_2,y_2)$. This means $x_1 \leq x_2$ or $y_1 \leq y_2$. For $R$ to be symmetric, this must imply $x_2 \leq x_1$ or $y_2 \leq y_1$. Let's find a counterexample. Consider the pairs $(1,1)$ and $(2,2)$. Is $(1,1)\,R\,(2,2)$? We check if $1 \leq 2$ or $1 \leq 2$. Since $1 \leq 2$ is true, $(1,1)\,R\,(2,2)$ holds. Now, let's check if $(2,2)\,R\,(1,1)$. We check if $2 \leq 1$ or $2 \leq 1$. Both $2 \leq 1$ are false. Therefore, $(2,2)\,R\,(1,1)$ does not hold. Since $(1,1)\,R\,(2,2)$ holds but $(2,2)\,R\,(1,1)$ does not hold, the relation $R$ is not symmetric. Based on Step 2 and Step 3, statement (I) "$R$ is reflexive but not symmetric" is correct. Step 4: Check for Transitivity A relation $R$ is transitive if for every $a, b, c$ in its domain, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$. Here, we need to check if for any $(x_1,y_1), (x_2,y_2), (x_3,y_3) \in \mathbb{N}\times\mathbb{N}$, if $(x_1,y_1)\,R\,(x_2,y_2)$ and $(x_2,y_2)\,R\,(x_3,y_3)$, then $(x_1,y_1)\,R\,(x_3,y_3)$. This means: If ($x_1 \leq x_2$ or $y_1 \leq y_2$) AND ($x_2 \leq x_3$ or $y_2 \leq y_3$), then it must follow that ($x_1 \leq x_3$ or $y_1 \leq y_3$). Let's try to find a counterexample. We want to find $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ such that: 1. $(x_1,y_1)\,R\,(x_2,y_2)$ is true. 2. $(x_2,y_2)\,R\,(x_3,y_3)$ is true. 3. $(x_1,y_1)\,R\,(x_3,y_3)$ is false. For $(x_1,y_1)\,R\,(x_3,y_3)$ to be false, we must have BOTH $x_1 > x_3$ AND $y_1 > y_3$. Let's choose $(x_1,y_1) = (3,3)$ and $(x_3,y_3) = (1,1)$. Then $(3,3)\,R\,(1,1)$ is false because $3 \not\leq 1$ and $3 \not\leq 1$. Now we need to find an $(x_2,y_2)$ such that: a) $(3,3)\,R\,(x_2,y_2)$ is true $\implies 3 \leq x_2$ or $3 \leq y_2$. b) $(x_2,y_2)\,R\,(1,1)$ is true $\implies x_2 \leq 1$ or $y_2 \leq 1$. Let's choose $(x_2,y_2) = (3,1)$. Check condition (a): Is $(3,3)\,R\,(3,1)$? This means $3 \leq 3$ or $3 \leq 1$. Since $3 \leq 3$ is true, $(3,3)\,R\,(3,1)$ holds. Check condition (b): Is $(3,1)\,R\,(1,1)$? This means $3 \leq 1$ or $1 \leq 1$. Since $1 \leq 1$ is true, $(3,1)\,R\,(1,1)$ holds. So, we have $(3,3)\,R\,(3,1)$ and $(3,1)\,R\,(1,1)$, but we found earlier that $(3,3)\,R\,(1,1)$ is false. Thus, the relation $R$ is not transitive. Based on this, statement (II) "$R$ is transitive" is incorrect. Step 5: Conclude based on the statements From Step 3, statement (I) ($R$ is reflexive but not symmetric) is correct. From Step 4, statement (II) ($R$ is transitive) is incorrect. Therefore, only statement (I) is correct. The final answer is $\boxed{\text{Only (I) is correct.}}$