# Continuous functions and infinum

Let $f:\mathbb R \to \mathbb R$ with $f(-2)=4$ and $f(3)=7$. Let $S:=\{x \in [-2,3]\mid f(x)\geq 5\}$. Then $\alpha:=\inf S$ exists. If $f$ is continuous at $\alpha$, show that:

(a) $-2<\alpha<3$

(b) $f(\alpha)=5$

To solve the question, I guess we should use the IVT which states: Let $f$ be continuous on $[a,b]$. If $k$ is any number between $f(a)$ and $f(b)$, then there exists $t \in (a,b)$ such that $f(t)=k$.

Or should we use the max/min theorem?

I'm not exactly sure which one to use and how to apply it here. Could anyone help me out please? Thanks.

• There is no typo, since changing $[-2,3]$ to $(-2,3]$ will not alter which points lie in $S$. – Cameron Buie Dec 13 '14 at 0:13
• But wouldn't $(-2,3]$ imply that $-2$ is not included in $S$? And since it's mentioned $f(-2)=4$ but in the set it says $f(x) \geq 5$? – Su003 Dec 13 '14 at 0:18
• True, $(-2,3]$ would imply that $-2\notin S.$ However, $f(-2)=4\not\ge 5$ does that, too. Basically, it isn't necessary to specify that $-2\notin S$ twice. Moreover, specifying that $-2\notin S$ does not automatically rule out the possibility that $\alpha\ne-2$--for example, we might have $S=(-2,3]$! – Cameron Buie Dec 13 '14 at 0:36
• ok, I'll edit the question then. Thanks. – Su003 Dec 13 '14 at 0:39

Hint: use the greatest-lower-bound property for $S$, say:

If $E\subset X$, $E$ is not empty, and $E$ is bounded below, then $\inf{E}\in X$

Take $E=S$, $X=[2,3]$, how $3\in S\implies S\neq \emptyset$, and $2$ is a lower bound of $S$, then $\inf{S}=\alpha\in[2,3]$.

Now for (a)

(i) If $\alpha=2$, For all $B_\delta(2)$, there is $x_0\in S$ and $x_0\in B_\delta(2)$ because $2=\inf{S}$, Take $\epsilon<1$, then by continuity $$|x_0-2|<\delta\implies|f(x_0)-f(2)|=|f(x_0)-4|<1$$ but how $f(x_0)\geq5\implies|f(x_0)-4|\geq 1$ this is a contradiction.

(ii) If $\alpha=3$, how $3=\inf{S}$, by definition of $S$, if $x_0\in[2,3)\implies f(x_0)<5$, take $\epsilon<2$, then by continuity $$|x_0-3|<\delta\implies|f(x_0)-f(3)|=|f(x_0)-7|<2$$ but how $f(x_0)<5\implies|f(x_0)-7|> 2$ this is a contradiction.

Therefor $2<\alpha<3$

For (b) Use (a), suposse that $f(\alpha)=t\neq5$, then if $t<5$ use (i), if $t>5$ use (ii).

The infimum (plural infima) of a subset S of a partially ordered set T is the greatest element of T that is less than or equal to all elements of S.

Thus, the infimum of your subset $S$ of the poset $\mathbb{R}$ (more specifically a chain) is the greatest element $x \in \mathbb{R} : f(x) \geq 5$ that is less than or equal to all elements of $S$.

Now consider the fact that $\alpha = inf \space S$ may not actually lie in $S$. So then it lies outside, say $x \leq -2$ or $3 \leq x$. This implies that all of the $f(x) : -2 < x < 3$ are less than $5$. But that's impossible by applying the IVT to $f(3)$ and any point to the left of $3$.

$b)$ follows from $a)$.

• What is a poset? – Su003 Dec 13 '14 at 0:25
• A poset is a partially ordered set having the following axioms: $a ≤ a$ (reflexivity); if $a ≤ b$ and $b ≤ a$ then $a = b$ (antisymmetry); if $a ≤ b$ and $b ≤ c$ then $a ≤ c$ (transitivity). In this case we used the operation "less than or equal to". But if we used the operation "<" then it would not be a poset. Hence, the reals under the operation $\leq$ is a poset. – user198428 Dec 13 '14 at 0:26
• Also, what would happen when we apply the IVT on $f(3)$ and any point to the left of $3$? Could you please clarify that part for me? Thanks – Su003 Dec 13 '14 at 0:33
• Take a variable $\psi = 3 - \epsilon$, where $\epsilon$ is really small so that $\psi$ is immediately next to $3$. Then take $f(\psi)$. Two cases appear: $f(\psi) \geq 5$, in which case you are done. Otherwise, $f(\psi) < 5$, suppose for example $f(\psi) = 3$. But connect a line between $f(\psi)$ and $f(3)$. What do you see? – user198428 Dec 13 '14 at 0:36

a) Assume that $\alpha=-2$. Then $f(x)\geq5$ for each $x>-2$. But then $f$ cannot be continuous at $\alpha$, since $f(\alpha)=f(-2)=4$. Assume that $\alpha=3$. Then $f(x)<5$ for each $x<3$. But then $f$ cannot be continuous at $\alpha$, since $f(\alpha)=f(3)=7$.

b) Since $\alpha$ is the infimum of $S$ we find sequences $(x_n)\subset S$, $(y_n)\subset[-2,3]\backslash S$ converging to $\alpha$. Note that $(a)$ implies that $[-2,3]\backslash S$ is not empty. Since $f$ is continuous at $\alpha$ we have $f(\alpha)=f(\lim_{n\to\infty}x_n)=\lim_{n\to\infty}f(x_n)\geq5\geq\lim_{n\to\infty}f(y_n)=f(\lim_{n\to\infty}y_n)=f(\alpha)$, so $f(\alpha)=5$.