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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 $x$ by contradiction.

Proof: Suppose $f$ is not continuous at $a$. If $f$ is not continuous at $a$ then $a$ cannot be an isolated point, since every function is continuous at an isolated point of its domain. If $f$ is not continuous there is some epsilon, for which no matter how what $\delta$ we choose there is a point $x_n \in S$ with $ \lVert f(x_n) − f(a)\rVert \geq \epsilon$. So let’s take $\delta = 1/n$ and $x_n \in S, \lVert x_n − a\rVert < 1/n$, $\lVert f(x_n) − f(a)\rVert \geq \epsilon$. Then $x_n \to a$ but $f(x_n) \to f(a)$. Hence some sequence of points $x_n$ converges to $a$ but $f(x_n)$ does not converge to $f(a)$

So a few questions for this. Can I use any arbitrary letter instead of epsilon? like $M$?

Also can anyone explain in better detail the part after If $f$ is not continuous there is some $\epsilon$. I am confused from that point on.

egreg
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RockChalk
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  • Please try to use MathJax(LaTeX) in future posts. I have edited your question. Please make sure that I didn't change the meaning of what you are asking. – Thomas Oct 16 '17 at 18:31
  • In principle, you can use whatever letter you want, as long as you don't use the same letter for two different things. In practice, your work will be easier to read if you use standard notation (like $\varepsilon$ and $\delta$). It would also be better to use the same notation in your answer as in the question, rather than changing $x$ to $a$ and changing $p_n$ to $x_n$. – Andreas Blass Oct 16 '17 at 20:56

1 Answers1

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Your idea is good. Indeed $x$ cannot be an isolated point for the reason you state.

Suppose $x$ is an accumulation point of the domain and that $f$ is not continuous at $x$. Then there exists $\varepsilon>0$ such that, for every $\delta>0$, there exists $p$ with $\|p-x\|<\delta$ and $\|f(p)-f(x)\|\ge\varepsilon$.

(How you call $\varepsilon$ is irrelevant.)

Now we can apply the condition with $\delta=1/n$, where $n$ is a positive integer, getting a sequence $(p_n)$ with $\|p_n-x\|<1/n$ and $\|f(p_n)-f(x)\|\ge\varepsilon$.

What you need to show now is that the sequence $(p_n)$ converges to $x$, which is true because $\lim_{n\to\infty}\|p_n-x\|=0$ by squeezing.

This is the required contradiction: indeed, the hypothesis tells you that the sequence $(f(p_n))$ converges to $f(x)$ hence there exists $N$ such that, for $n>N$, $\|f(p_n)-f(x)\|<\varepsilon$ (the same $\varepsilon$ as before!).

However, by construction, $\|f(p_n)-f(x)\|\ge\varepsilon$ for all $n$.

egreg
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