We show that the [[Image of a set under a function|image]] of an interval under a continuous function is an Interval. **Lemma** Let $I$ be an [[Real intervals]] and suppose that $f: I \to R$ is a [[Continuous Real Function|continuous]]. Then $f(I)$ is an interval. **Proof** Choose $a,b\in f(I)$ with $a<b$ then there exists $x,y \in I \;(x \neq y)$ such that $f(x) = a$ and $f(y)=b$. WTS $[a,b] \subset f(I)$ (using interval definition). Suppose that $x < y$. Then for any $g\in (a,b)$ there exists a point $c \in (x,y)$ such that $f(c)=g$ by [[Intermediate Value Theorem (IVT)|IVT]] so $g\in f(I)$. It follows that $[a,b] \subset f(I)$. The other three cases can be treated similarly. **Remark** - Using [[Intermediate Value Theorem (IVT)|EVT]], a corollary of this is that the [[Image of Closed Real Interval under Continuous Real Function is Closed Real Interval]]. - On the other hand, [[Continuous image of an open interval can be any kind of interval]].