Image of Interval by Continuous Function is Interval

Theorem
Let $$f$$ be a real function which is continuous on an interval $$I$$.

Then the image of $$I$$ by $$f$$ is also an interval.

Proof
Let $$J = f \left({I}\right) = \left\{{f \left({x}\right): x \in I}\right\}$$.

From the property defining an interval, we need to show that if $$J$$ is an interval, then $$y_1, y_2 \in J, y_1 \le \lambda \le y_2 \implies \lambda \in J$$.

So suppose $$y_1, y_2 \in J$$, and suppose $$y_1 \le \lambda \le y_2$$.

Consider these subsets of $$I$$:
 * $$S = \left\{{x: f \left({x}\right) \le \lambda}\right\}$$
 * $$T = \left\{{x: f \left({x}\right) \ge \lambda}\right\}$$

As $$y_1 \le \lambda \in J$$ and $$y_2 \ge \lambda \in J$$, it follows that $$S$$ and $$T$$ are both non-empty.

Also, every point of $$I$$ belongs to either $$S$$ or $$T$$.

So from Interval Divided into Subsets, a point in one subset is at zero distance from the other.

So, suppose that $$s \in S$$ is at zero distance from $$T$$.

From Limit of Sequence to Zero Distance Point, we can find a sequence $$\left \langle {t_n} \right \rangle$$ in $$T$$ such that $$\lim_{n \to \infty} t_n = s$$.

Since $$f$$ is continuous on $$I$$, it follows from Limit of Image of Sequence that $$\lim_{n \to \infty} f \left({t_n}\right) = f \left({s}\right)$$.

But $$\forall n \in \N^*: f \left({t_n}\right) \ge \lambda$$.

Therefore by Lower and Upper Bounds for Sequences, $$f \left({s}\right) \ge \lambda$$.

We already have that $$f \left({s}\right) \le \lambda$$.

Therefore $$f \left({s}\right) = \lambda$$ and so $$\lambda \in J$$.

A similar argument applies if a point of $$T$$ is at zero distance from $$S$$.