Image of Interval by Continuous Function is Interval/Proof 1

Proof
Let $J$ be the image of $f$.

By definition of real interval, it suffices to show that:
 * $\forall y_1, y_2 \in J: \forall \lambda \in \R: y_1 \le \lambda \le y_2 \implies \lambda \in J$

So suppose $y_1, y_2 \in J$, and suppose $\lambda \in \R$ is such that $y_1 \le \lambda \le y_2$.

Consider these subsets of $I$:
 * $S = \set {x \in I: \map f x \le \lambda}$
 * $T = \set {x \in I: \map f x \ge \lambda}$

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

Also, $I = S \cup T$.

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 $\sequence {t_n}$ in $T$ such that $\ds \lim_{n \mathop \to \infty} t_n = s$.

Since $f$ is continuous on $I$, it follows from Limit of Image of Sequence that:
 * $\ds \lim_{n \mathop \to \infty} \map f {t_n} = \map f s$

But:
 * $\forall n \in \N_{> 0}: \map f {t_n} \ge \lambda$

Therefore by Lower and Upper Bounds for Sequences:
 * $\map f s \ge \lambda$

We already have that:
 * $\map f s \le \lambda$

Therefore:
 * $\map f s = \lambda$

and so:
 * $\lambda \in J$

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