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Here is the theorem in math:

A real value function $f$ is continuous at $x \in \Bbb R$ iff whenever a sequence of real numbers $x_{n}$ converges to $x$ then the sequence $f(x_{n})\to f(x)$.

For proving the converse, it seems easiest to do a proof BWOC. I am wondering if a direct proof is possible, and if so whether this proof works:

PF: Assume that whenever a sequence of real numbers $x_{n}$ converges to $x$ then the sequence $f(x_{n}) \to f(x)$.

Then $\forall \epsilon > 0$ $\exists N$ s.t. $n\geq N$ implies $\lvert x_{n} - x \rvert < \epsilon$, which further implies that $\forall \epsilon^* > 0$ $\exists N^*$ s.t. $n^* \geq N^*$ implies $\lvert f(x_{n}) - f(x) \rvert < \epsilon^*$

Now it needs to be shown that $f$ is continuous at $x$ i.e. $\forall \epsilon_1 >0 $ $\exists \delta >0$ s.t $0<\lvert x_n - x \rvert < \delta $ implies $\lvert f(x_n) - f(x) \rvert < \epsilon_1$ (sorry for reusing $x_n$)

This definition for continuous seems to be almost the same as the definition of $f(x_{n})$ $\rightarrow f(x)$? If so, can't I just construct a sequence, say {$ x_n \lvert x_n = x+ \frac{1}{n}$} and then choose $\delta$ such that $\delta = x_N - x $ where $x_N$ is what makes $\forall \epsilon > 0$ $\exists N$ s.t. $n\geq N$ implies $\lvert x_{n} - x \rvert < \epsilon$, which further implies that $\forall \epsilon^* > 0$ $\exists N^*$ s.t. $n^* \geq N^*$ implies $\lvert f(x_{n}) - f(x) \rvert < \epsilon^*$?

So.... does this concept work, or am I overlooking something?

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  • $\begingroup$ You got $\varepsilon$ and $\delta$ mixed up. Also, there is a flaw in your proof, namely - you want to show that $|f(x_n)-f(x)|<\varepsilon$ for $every$ $x_n$ in a $\delta$ neighborhood, and you showed it only for $x$'s of the form $x+1/n$. In truth, it is not entirely clear whether this concept will work. $\endgroup$ – uniquesolution Aug 31 '15 at 19:18
  • $\begingroup$ uniquesolution is quite right. $\endgroup$ – R.N Aug 31 '15 at 19:21
  • $\begingroup$ $x_n$ is arbitrary sequence so what is exactly $x_N$? $\endgroup$ – nakajuice Aug 31 '15 at 19:24
  • $\begingroup$ @user265678 - Even if you use all the sequences in the world the proof would not be so pleasant. You see, you have an arbitrary $x_0$ in a $\delta$ neighborhood of $x$. Now you must come up with a sequence that converges to $x$, that will somehow imply that the value of the function at $x_0$ is $\varepsilon$-close to $f(x)$. $\endgroup$ – uniquesolution Aug 31 '15 at 19:38
  • $\begingroup$ @uniquesolution I see. I will stick with proof by contradiction then. Thank you (sorry I deleted my first comment then was typing something else and saw that you had replied to my original comment) thank you very much. $\endgroup$ – majmun Aug 31 '15 at 19:38

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