Which of the following is an empty set?

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(A) For option $A$: $x^2+1=0 \Rightarrow x^2=-1$. Since the square of any real number $x \in \mathbb{R}$ is always non-negative $(x^2 \ge 0)$,there is

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$\left\{x: x^2+1=0, x \in \mathbb{R}\right\}$

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(A) For option $A$: $x^2+1=0 \Rightarrow x^2=-1$. Since the square of any real number $x \in \mathbb{R}$ is always non-negative $(x^2 \ge 0)$,there is no real number $x$ that satisfies $x^2=-1$. Thus,the set is empty.For option $B$: $x^2-9=0$ $\Rightarrow x^2=9$ $\Rightarrow x = \pm 3$. This set is $\{3, -3\}$,which is not empty.For option $C$: $x^2-x-2=0$ $\Rightarrow (x-2)(x+1)=0$ $\Rightarrow x=2, -1$. This set is $\{2, -1\}$,which is not empty.For option $D$: $x^2-1=0$ $\Rightarrow x^2=1$ $\Rightarrow x = \pm 1$. This set is $\{1, -1\}$,which is not empty.Therefore,the correct option is $A$.

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