2

Let f be a function defined at x. Suppose that every sequence $p_1, p_2, p_3 \dots$ in the domain of f converging to $x$ has the property that $f(p_1), f(p_2),f(p_3), \dots$ converges to $f(x)$. Prove that $f$ is continuous at $x$.

So I could prove this by contradiction. So I would start by assuming $f$ is not continuous at $x$. I know I need to create a sequence of points $p_1, p_2, p_3, \dots$ that converges to $x$, but $f(p_1), f(p_2), f(p_3), \dots$ does not converge to $f(x)$.

Confused on how to write this up formally and also steps I should be taking?

Matthew Leingang
  • 26,112
RockChalk
  • 69
  • This is the right idea, and dbx's answer is a good one. One technical point: this manner of proof is called proof by contrapositive, rather than contradiction. – Matthew Leingang Oct 09 '17 at 17:06

1 Answers1

1

Your hypothesis is sometimes taken as the definition of continuity; I'll assume then that the definition you want to use is:

For all $\varepsilon > 0$ there is a $\delta > 0$ such that $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$.

You should use the definition of a convergent sequence, i.e. $(a_n)$ is convergent with $a_n\rightarrow a$ iff for all $\varepsilon > 0$ there is an $N\in\Bbb{N}$ such that $n\geq N$ implies $|a_n-a|<\varepsilon$.

If $f$ is not continuous, then you can find an $\varepsilon>0$ such that for every $\delta>0$ there is a point $a$ in the domain with $|x-a|<\delta$ but $|f(x)-f(a)|\geq\varepsilon$. In particular you can choose $\delta = 1/n$, and find a point $a_n$ for each $n$. These points define a convergent sequence in the domain, but their images do not converge, a contradiction.

You'll have to fill in a lot of details, but that's the sketch.

Sort of Damocles
  • 2,443