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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?

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  • $\begingroup$ 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. $\endgroup$ – Offlaw Nov 26 '18 at 13:24
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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.

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  • $\begingroup$ “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. $\endgroup$ – Offlaw Nov 26 '18 at 13:30
  • $\begingroup$ "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 $\endgroup$ – Aishwarya Deore Nov 26 '18 at 13:42
  • $\begingroup$ @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? $\endgroup$ – José Carlos Santos Nov 26 '18 at 17:25
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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.

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