Is 2 <= x <= 1 a predicate or a statement?

Question

I can understand that a predicate is a predicate because the truth or falsity of it depends on the specific value of the 'variable' or the 'part acting as a variable' in it. (specific value should be from the domain of course).

But let's have this predicate: 2 <= x <= 1 (Domain is all real numbers)

Now whatever value we substitute for x, the predicate results in a statement which is false and hence there is absolutely no doubt about its truth value. It is false independent of the value replaced.

Does this make this predicate a statement essentially?

Answer 1

I'd caution against defining a predicate as having a truth value "depending on" the arguments; one should say it's determined by them, which is consistent with all choices given the same result. It's a bit like functions. You presumably want the sum of two $\mathbb{R}\mapsto \mathbb{R}$ functions to also be a function, but $x^3,\,1-x^3$ have a constant sum. Similarly, you'd want the conjunction of $2\le x,\,x\le 1$, with $x$ fixed (variable), to be a statement (predicate), just like the conjuncts themselves are. So even through $2\le x\le 1$ is identically false, it's still a statement if $x$ is specified, or a unary predicate if it isn't.

Answer 2

A statement (sentence) does not have any free variables. So in isolation, $2 \le x \le 1$ is not a sentence, for syntactical reasons. If you add a quantifier it becomes a sentence: $\forall x: 2 \le x \le 1$. This is a false sentence in the model where the domain is all real numbers and the symbols have their usual meanings. Alternatively you can add an existential quantifier: $\exists x: 2 \le x \le 1$. This is obviously also a false sentence.

Viewed as a predicate, $2 \le x \le 1$ is a perfectly good predicate since it has $x$ as a parameter. Compare with a function $f(x)$ where a special case is $f(x)=0$ for all values of $x$. Here $f$ is still a function rather than a constant symbol, for syntactical reasons.