# Continuity of functions of monotone sequences

Suppose $$\lim_{n\to\infty} f(x_n) = f(c)$$ for any monotone sequence $$x_n$$ approaching $$c$$. The prove that $$f$$ is continuous at $$c$$.

Solution: We prove it by contradiction. Assume $$f$$ is not continuous at $$c$$. Then there exists $$\varepsilon > 0$$ such that for any $$n$$ belonging to $$\mathbb{N}$$, there is $$x_n$$ such that $$x_n$$ approaches c but $$|f(x_n) - f(c)| > \varepsilon$$. Then there is subsequence $$x_{n_k}$$ such that $$\lim_{k\to\infty} x_{n_k} = c$$ and $$x_{n_k}$$ is monotone. Then by assumption we have $$\lim f(x_{n_k}) = f(c)$$ which is a contradiction.

Is this right? Can someone explain why we have a contradiction?

• If $f(x_n) \to f(c)$ for any sequence $\{x_n\}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered. – Offlaw Nov 26 '18 at 13:24

## 2 Answers

No, it is not right. The assertion “there is $$x_n$$ such that $$x_n$$ approaches $$c$$ but $$\bigl|f(x_n)-f(c)\bigr|>\varepsilon$$” makes no sense.

There is a $$\varepsilon>0$$ such that, for each $$\delta>0$$, there is a $$x\in(c-\delta,c+\delta)\cap D_f$$ such that $$\bigl\lvert f(x)-f(c)\bigr\rvert\geqslant\varepsilon$$. In particular, for any natural $$n$$, there is a $$x_n\in\left(c-\frac1n,c+\frac1n\right)\cap D_f$$ such that $$\bigl\lvert f(x_n)-f(c)\bigr\rvert\geqslant\varepsilon$$. The sequence $$(x_n)_{n\in\mathbb N}$$ has a monotonic subsequence $$(x_{n_k})_{k\in\mathbb N}$$ and, since $$\lim_{n\to\infty}x_n=c$$, $$\lim_{k\to\infty}x_{n_k}=c$$. Therefore, we should have $$\lim_{k\to\infty}f(x_{n_k})=f(c)$$. But we don't, since $$(\forall k\in\mathbb{N}):\bigl\lvert f(x_{n_k})-f(c)\bigr\rvert\geqslant\varepsilon$$. So, we have a contradiction here.

• “there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>\epsilon$" - Why does this not make sense? Please I can't understand. – Offlaw Nov 26 '18 at 13:30
• "x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε – Aishwarya Deore Nov 26 '18 at 13:42
• @AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $\mathbb N$, there is $x_n$” mean? – José Carlos Santos Nov 26 '18 at 17:25

No. Mostly, your third sentence is mangled. Here's a corrected version:

If $$f$$ is not left continuous at $$c$$, then there is some $$\varepsilon > 0$$ such that for any $$\delta > 0$$, there is some $$x \in (c-\delta,c)$$ such that $$|f(x)-f(c)| \geq \varepsilon$$. Choosing, in particular, for each $$n \in \mathbb{N}$$, $$\delta = \frac{1}{n}$$, and choosing some $$x_n \in (c-\delta,c)$$, we obtain a sequence $$(x_n)$$ such that $$(x_n)\to c$$ but $$|f(x_n)-f(c)| \geq \varepsilon$$ for all $$n$$. Now, every sequence has a monotone subsequence, so in particular $$(x_n)$$ has a monotone subsequence $$(x_{n_k})$$, and $$(x_{n_k})\to c$$, since it's a subsequence of $$(x_n)$$, and $$|f(x_{n_k})-f(c)|\geq\varepsilon$$ for all $$k$$. This contradicts our hypothesis about $$f$$, so $$f$$ is left continuous.

An identical proof (with $$(c-\delta,c)$$ replaced by $$(c,c+\delta)$$) shows that $$f$$ is right-continuous [or you could just patch this into the main proof if you prefer], so $$f$$ is both left- and right-continuous, so is continuous.