Suppose that the function $f:I\rightarrow\mathbb{R}$ is monotonically increasing and bounded. Prove that the $\lim_{x\rightarrow a}f(x)$ exists

Question

Let $a$ and $b$ be numbers with $a<b$ and set $I=(a,b)$. Suppose that the function $f:I\rightarrow\mathbb{R}$ is monotonically increasing and bounded. Prove that the $\lim_{x\rightarrow a}f(x)$ exists.

Observe that $a=\inf(I)$ where $a\notin(a,b)$, and since $f(I)$ is bounded, there exists an $\alpha=\inf f(I)$. Now, let $\{x_n\}$ be sequence of $I$ that converges to $a$, then there exists a subsequence $\{x_{n_k}\}$ that also converges to $a$. Since $f:I\rightarrow\mathbb{R}$ is monotonically increasing and bounded, we have $f(\{x_{n_k}\})$ is monotone bounded decreasing that converges to $\alpha$ by the monotone convergent theorem. Now, let $\epsilon>0$, there exists a $K\in\mathbb{N}$ such that $\alpha\leq f(x_{n_k})<\alpha+\epsilon$. Since $\{x_n\}$ converges to $a$, there exists a $N\in\mathbb{N}$ such that $a\leq\{x_n\}<\{x_{n_k}\}$ for all $n\geq N$. Then we have $\alpha\leq f(x_n)\leq f(x_{n_k})<\alpha+\epsilon$; thus we $\lim_{x\rightarrow a}f(x)=\alpha$

Can someone check this solution right or not? I am still struggling with sequences. Thanks in advanced.

What do you mean by $I\not\in(a,b)$? You have defined $I=(a,b)$? Also, the limit should be $\lim_{x\rightarrow a^+}f(x)$? — charlestoncrabb, Nov 08 '15 at 00:20
You forget to mention that your subsequence is decreasing. (You could have instead just claimed your original sequence ${x_n}$ was chosen to be decreasing which would save you considerable grief.) — B. S. Thomson, Nov 08 '15 at 00:57

score 1 · Accepted Answer · answered Nov 08 '15 at 01:09

You are not wrong, but in quoting the monotone convergence theorem, you may actually simplify the proof a lot.

Pick some $x_n\rightarrow a^+$, and WLOG suppose the sequence is strictly decreasing (there is no need to pass to a subsequence). By monotonicity and boundedness of $f$, the sequence $f(x_n)$ is decreasing (at least non-increasing), hence converges by monotone convergence theorem. We are actually done at this point since the sequence $x_n$ was chosen arbitrarily.

Short answer: you are not wrong, though a little bit of confusion is apparent when you make the proof more complicated than it needs to be.

score 1 · Answer 2 · answered Nov 08 '15 at 01:14

Given the facts we may only show that the limit from the right exists.

Let $\varepsilon>0$ be given. Since $f$ is bounded on $(a,b)$, $\, \inf_{x \in (a,b)} f(x)$ exists. Letting $L = \inf_{x \in (a,b)} f(x)$ we may find $x_0 \in (a,b)$ so that $L \leq f(x_0) < L + \varepsilon$. Set $\delta=x_0-a$. Since $f$ is increasing we have that, if $a < x < a + \delta$ then $L \leq f(x) < L + \varepsilon$.

Suppose that the function $f:I\rightarrow\mathbb{R}$ is monotonically increasing and bounded. Prove that the $\lim_{x\rightarrow a}f(x)$ exists

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