Closed Point Set and Continuity If $f$ is a continuous function whose domain includes the closed interval [$a,b$] and there is a point $x$ in [$a,b$] so that $f(x)$ is greater than or equal to zero, then $\{ x \in [a,b] \mid f(x) \geq 0\}$ is a closed set.
Proof:
Let $f$ be a continuous function whose domain includes the closed interval [$a,b$]
Let $x$ be a point so that $f(x) \geq 0$.
Then 


*

*$p$ is a point on $f$, and

*if $S$ is any open interval containing the number $f(x$), then there is an open interval $T$ containing the number $x$ such that if $t ∈ T$, and $t$ is in the domain of $f$, then $f(t) ∈ S$. (Our class' definition of continuous)


(The statement that the point set $M$ is a closed point set
means that if $p$ is a limit point of $M$, then $p$ is in $M.$
Note that if a set $M$ has no limit point, then it is a closed point set. We could equivalently define closed by saying that $M$ is closed if, and only if, there is no limit point of $M$ that is not in $M$. (This is our class' definition of closed, however I am unsure on where to go from here.))
(Also here is our class' definition of a limit point: If $M$ is a point set and $p$ is a point, the statement that $p$ is a limit point of the point set $M$ means that every open interval containing $p$ contains a point of $M$ different from $p$.)
Any help on where to go would be appreciated.
 A: Your definition of continuous can be rephrased as "the preimage of an open set is open".  (Let $f:X \rightarrow Y$.  The preimage of a set, $S \subset Y$ is the set $\{x \in X : f(x) \in S\}$.  That is, it is all points of $X$ that $f$ takes to some point in $S$.)
Hint:  For your question, the interesting sets in the codomain are $U = (-\infty,0)$ and $V = [0,\infty)$, which are clearly complementary.  Continuity of $f$ tells you that one of those has an open preimage.  Your data about $f(x)$ tells you that the other is nonempty.  What can you say about the (nonempty) complement of an open set?
(Actually, nonemptiness is not needed.  Since $\varnothing$ is both open and closed, even if the complement is empty, it is still closed.)
A: This is a matter of definition of closed set and basic properties of a continuous function. Consider the set $$S=\{x\mid x\in[a, b], f(x) \geq 0\} $$ It is given that $S$ is non-empty. Let $p$ be a limit point of $S$ and our job will be done if we show that $p\in S$. Since $S\subseteq [a, b] $ and $[a, b] $ is a closed interval the point $p\in[a, b] $. On the contrary let's assume that $p\notin S$. Therefore $f(p) <0$. By continuity there is a neighborhood $I$ of $p$ such that $f(x) <0$ for all $x\in I\cap[a, b] $. Hence no point of $I$ is in $S$. This contradicts the fact that $p$ is a limit point of $S$. Therefore $p\in S$. 
