5. Continuity

Let $$ be irrational and let $$. Then for all $$, there exists $$ (by the Density theorem) and therefore $$. Hence, $$ is discontinuous at $$. A similar argument shows that $$ is discontinuous $$. Alternatively, if $$ then there exists a sequence of irrational numbers $$ converging to $$. Now $$ and $$, and this proves that $$ is discontinuous at $$. A similar arguments holds for $$ irrational.

Let $$ with $$. There exists a sequence $$ of irrational numbers in $$ converging to $$. Hence, $$ while $$. This shows that $$ is discontinuous at $$. Now let $$ be irrational and let $$ be arbitrary. Let $$ be such that $$. In the interval $$, there are only a finite number of rationals $$ with $$, otherwise we can create a sequence $$ with $$, all the rationals $$ distinct and thus necessarily $$ is unbounded. Hence, there exists $$ such that the interval $$ contains only rational numbers $$ with $$. Hence, if $$ then $$ and therefore $$. On the other hand, if $$ is irrational then $$. This proves that $$ is continuous at $$.

Let $$ be arbitrary. By continuity of $$ and $$ at $$, there exists $$ such that $$ for all $$ such that $$, and there exists $$ such that $$ for all $$ such that $$. Let $$. Then for $$ such that $$ we have that $$ This proves that $$ is continuous at $$. A similar proof holds for $$. Consider now the function $$. If $$ then $$ for all $$ and continuity is trivial. So assume that $$. Let $$ be arbitrary. Then there exists $$ such that if $$, $$, then $$. Therefore, for $$, $$, we have that $$ We now prove continuity of $$. Let $$ be any sequence in $$ converging to $$. Then $$ converges to $$ by continuity of $$ at $$, and $$ converges to $$ by continuity of $$ at $$. Hence the sequence $$ converges to $$. Hence, for every sequence $$ converging to $$, $$ converges to $$. This shows that $$ is continuous at $$.

Let $$ be arbitrary. Let $$. If $$ then $$.

Let $$ be arbitrary. Then there exists $$ such that $$ for all $$. Therefore, if $$ then $$ and therefore $$

We have that $$ Hence given $$ we choose $$. The proof that $$ is continuous follows from the fact that $$ and Lemma 5.2.6.

For $$, we must consider $$. Given $$ let $$. Then if $$ and $$ then $$. This shows that $$ is continuous at $$. Now suppose that $$. Then $$ Hence, given $$, suppose that $$. Then $$.

Follows from the inequality $$.

Let $$ be given. Let $$ and let $$. Then there exists $$ such that if $$ then $$. Now since $$ is continuous at $$, there exists $$ such that if $$ then $$. Therefore, if $$ then $$ and therefore $$. This proves that $$ is continuous at $$. Since $$ is arbitrary, $$ is continuous on $$.

Suppose that $$ is unbounded. Then for each $$ there exists $$ such that $$. Now the sequence $$ is bounded since $$. By the Bolzano-Weierstrass theorem, $$ has a convergent subsequence, say $$, whose limit $$ satisfies $$. Since $$ is continuous at $$ then $$ exists and equal to $$. This is a contradiction since $$ implies that $$ is unbounded.

Let $$ be the range of $$. By Theorem 5.3.3, $$ exists; set $$. By the definition of the supremum, for each $$ there exists $$ such that $$. In particular, for $$, there exists $$ such that $$. Then $$. The sequence $$ is bounded and thus has a convergent subsequence, say $$. Let $$. Clearly, $$. Since $$ is continuous at $$, we have that $$. Hence, $$ is a maximum point. A similar proof establishes that $$ has a minimum point on $$.

Let $$. The set $$ is non-empty because $$. Moreover, $$ is clearly bounded above. Let $$. By the definition of the supremum, there exists a sequence $$ in $$ such that $$. Since $$ it follows that $$. By definition of $$, $$ and since $$ is continuous at $$ we have that $$, and thus $$. Now let $$ be such that $$ and $$. Then $$ converges to $$ and $$ because $$. Therefore, since $$ we have that $$. Therefore, $$ and this proves that $$. This shows also that $$.

Let $$ be the distance traveled along the trail on the way up the mountain after $$ units of time, and let $$ be the distance remaining to travel along the trail on the way down the mountain after $$ units of time. Both $$ and $$ are defined on the same time interval, say $$ if $$ is measured in hours. If $$ is the length of the trail, then $$, $$, $$ and $$. Let $$. Then $$ and $$. Hence, there exists $$ such that $$. In other words, $$, and therefore $$ is the time of day when the hiker is at exactly the same point on the trail.

The function $$ is continuous on $$. We have that $$ and $$. Therefore, there exists $$ such that $$, i.e., $$ has a zero in the interval $$.

Since $$ achieves its maximum and minimum value on $$, there exists $$ such that $$ for all $$. Hence, $$. Assume for simplicity that $$. Then $$. Let $$ be such that $$. Then by the Intermediate Value Theorem, there exists $$ such that $$. Hence, $$, and this shows that $$. Therefore, $$.

We have that $$. Hence, for any $$ we let $$, and thus if $$ then $$.

We have that $$ Hence, for $$ let $$ and if $$ then $$.

We have that $$ Now, $$ implies that $$ and therefore $$ It follows that $$. Hence, given $$ we let $$, and if $$ then $$.

We prove the contrapositive, that is, we will prove that if $$ is not uniformly continuous on $$ then $$ is not continuous on $$. Suppose then that $$ is not uniformly continuous on $$. Then there exists $$ such that for $$, there exists $$ such that $$ but $$. Clearly, $$. Now, since $$, by the Bolzano-Weierstrass theorem there is a subsequence $$ of $$ that converges to a point $$. Now, $$ and therefore also $$. Hence, both $$ and $$ are sequences in $$ converging to $$ but $$. Hence $$ and $$ do not converge to the same limit and thus $$ is not continuous at $$. This completes the proof.

Consider $$ and $$. Clearly $$. On the other hand $$

By assumption, $$ for all $$ for some constant $$. Let $$ be arbitrary and let $$. Then if $$ then $$. Hence, $$ is uniformly continuous on $$.