Proof verification : $\lim_{x \to c^+}f(x)=\inf\{f(x): x>c\}$ for increasing $f$

Question

The Problem : $J \subset \mathbb{R}$ is an interval. Let $f : J \to \mathbb{R}$ be an increasing function. Let $c \in J$ such that $c$ is not the right-end point of $J$ $($i.e. $c < \sup(J))$. To show that $$\lim_{x \to c^+}f(x)=\inf\{f(x): x \in J; x>c\}$$ Where $\lim_{x\to c^+}f$ represents the one-sided right-hand limit, which is clarified at the start of my solution.

I'm attaching my solution. Please notify if there's any gap/flaws in the arguments, also whether it could be made shorter by any other technique. Any comment/suggestion regarding this proof, or maybe in general context (style of proof-writing etc), would be greatly appreciated. Thank you.

My Solution : Let $\lim_{x \to c^+}f(x)=l$ i.e. for every $\epsilon>0, \exists \delta=\delta(\epsilon)>0$ such that $x \in (c, c+\delta)\cap J \implies l-\epsilon<f(x)<l+\epsilon$

Let $E=\{f(x) : x \in (c, \infty) \cap J \}$ and $E_{\delta}=\{f(x) : x \in (c, c+\delta) \cap J \}.$ We are required to show that $\inf(E)=l.$

Clearly, $E_{\delta} \subseteq E.$ Hence, $\inf(E_{\delta}) \geq \inf(E).$ Now suppose $\inf(E_{\delta}) > \inf(E).$ We wish to contradict this statement.

Let $y_1 \in E_{\delta}, ~y_2 \in E \setminus E_{\delta}.$ Then $\exists x_1 \in (c,c+\delta) \cap J$ such that $f(x_1)=y_1$ and $\exists x_2 \in [c+\delta, \infty) \cap J$ such that $f(x_2)=y_2$. Since $x_1 < x_2,$ and $f$ is increasing, we must have $y_1 \leq y_2.$ Note that $y_1 \in E_{\delta}$ and $y_2 \in E \setminus E_{\delta}$ is chosen arbitrarily. Hence any element of $E_{\delta}$ is less than or equal to any element of $E \setminus E_{\delta}.$

Let $\alpha = \frac{\inf(E_{\delta})-\inf(E)}{2}.$ Then $\inf(E)+\alpha$ is not a lower bound of $E$. Hence $\exists x \in (c,\infty) \cap J$ such that $f(x)=\inf(E)+\alpha.$

Now, $\inf(E)+\alpha < \inf(E_{\delta}) \Leftrightarrow f(x) < \inf(E_{\delta}) \Rightarrow f(x) \notin E_{\delta} \Rightarrow x \notin (c,c+\delta) \cap J$.

$[$The last step is the contrapositive of the following statement : $x \in (c,c+\delta) \cap J \Rightarrow f(x) \in E_{\delta},$ which is true by the definition of $E_{\delta}]$

Hence we must have $x \in [c+\delta, \infty) \cap J$. Let $z \in (c,c+\delta) \cap J$. Clearly $x>z,$ and since $f$ is increasing, $f(x) \geq f(z).$ Again, $z \in (c,c+\delta) \cap J \Rightarrow f(z) \in E_{\delta},$ and hence $\inf(E_{\delta}) \leq f(z).$ Coupled with the fact that $f(x) < \inf(E_{\delta}),$ we have $f(x)<f(z).$ A contradiction. Hence we must have $\inf(E_{\delta}) = \inf(E).$ Since $\epsilon > 0$ is arbitrary, we have

$$\inf(E_{\delta(\epsilon)}) = \inf(E)\quad\forall~ \epsilon > 0\tag{1}$$

We know that $x \in (c, c+\delta)\cap J \Rightarrow l-\epsilon<f(x)<l+\epsilon$. Then from the definition of $E_{\delta},$ $l-\epsilon \leq \inf(E_{\delta}) < l+\epsilon.$ Hence, by $(1), l-\epsilon \leq \inf(E) < l+\epsilon.$ Since $\epsilon > 0$ is arbitrary, we conclude that $\inf(E)=l$ (since otherwise, we can choose $\epsilon = \frac{|\inf(E)-l|}{2}$ and obtain a contradiction), which is what we set out to prove.

Answer 1

For all $x,y \in J$ such that $c < x < y$ we have $\inf(E) \le f(x)\leq f(y)$, Since $f$ is increasing.

Now by letting $x \rightarrow c^+$ in each sides of the inequality (while $y$ is fixed), we get $\inf(E) \leq l\leq f(y)$ for all $c<y \in J. $ Hence by taking infimum over $y: ~c<y$ from each side of the later inequality we get $\inf(E) \leq l \leq \inf(E)$.

Edit In case if the existence of $l$ is questionable:

Let $l_1 =\liminf_{x \to c^+} f(x)$ and $ l_2 =\limsup_{x \to c^+} f(x)$ then observe that $$\inf(E) \leq l_1 \leq l_2 \leq f(y).$$

for all $c < y$. Now take infimumu over all $y$ such that $c<y.$ Done

Answer 2

Let $c<x_n <\sup (J)$, $n=1,2, \dots$ such that $x_n\downarrow c$.

Note that $f[c,x_1]\supset f[c,x_2]\supset \dots \supset \lim_{n\to\infty}f[c, x_n]=A \dots (1)$.

Since $f(x)$ increasing function, $\sup\{f[c,x_1]\}\geq \sup\{f[c,x_2]\}\geq \dots \geq \sup \{A\}$. But $\sup \{A\} =\inf \{f(x):x\in J;x>c\}(*)$. On the other hand $\sup \{A\}= \lim_{x\rightarrow c^+} f(x)$

To prove (*), note that, (Assuming $A$ is non-empty):Note that,$\sup \{A\}\leq\inf \{f(x):x\in J;x>c\}$ trivial since we can always choose $M\in \mathbb{N}$ for which $x_*\geq x_M>c$ such that $\sup f[c,x_M]\leq \inf E(x_*)$ for every $x_*>c$ where $E(x_*)=\inf \{f(x):x\in J;x\geq x_* >c\}$, that gives us $\sup \{A \} \leq \inf\{ E(x_*)\}$ for all $x_*>c$ for $(1)$.

On the other hand Let $x_1>y>\sup \{A\}$, implies $\exists N$, such that $f(x_N)<y<f(x_{N+1})$, then y can not be $\inf \{f(x):x\in J;x>c\}$.Then $\sup \{A\}\geq\inf \{f(x):x\in J;x>c\}$ Again let $z<\inf \{f(x):x\in J;x>c\}$, then $z$ can not be equal to $\sup \{A\}$. Then $\sup \{A\}\geq\inf \{f(x):x\in J;x>c\}$. Hence $(*)$ holds.