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In the definition of continuity the interval $I$ may be replaced by an arbitrary finite union of intervals. Consider $X = (-1, 0) \cup (0, 1)$ and the function $f:X \to \mathbb{R}$ defined by $$ f(x)=\begin{cases} 1 &\text{if $x \in (0, 1)$}\\ -1 &\text{if $x \in (-1, 0)$} \end{cases} $$ Decide whether or not the function f is continuous and prove it using epsilon-delta.

My initial thoughts were that this was continuous by allowing epsilon = delta, though now I'm not sure...

Edit:

Thank you very much for editing it! I'm still quite new to this math jax business

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  • $\begingroup$ For $\varepsilon <2$ your $\delta $ must be small enough to keep the interval in the domain on one side of $0$. So your $\delta $ will need to depend on $x$. $\endgroup$
    – Ian
    Mar 15, 2016 at 15:50

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Hint. Let $c \in X$. Then either $c \in (-1,0)$ or $c \in (0,1)$. Then choose $\delta$ small enough to insure that $(c-\delta, c + \delta)$ does not contain $0$.

Hint 2. (After OP's comment). You are on the right track ! You have to prove that, for every $\varepsilon > 0$, there exists $\delta > 0$ (use Hint 1) such that $|x -c| < \delta$ implies $|f(x) - f(c)| < \varepsilon$. But as you observed, $|f(x) - f(c)| = 0 < \varepsilon$ as long as $x$ and $c$ are in the same interval. Can you conclude now?

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  • $\begingroup$ My original thought was that |f(x)-f(x0)| = 0 for all x both intervals which will always be smaller than any epsilon >0. My understanding of real analysis is very basic though and I admit I'm lost! $\endgroup$
    – Killuminati
    Mar 15, 2016 at 16:23
  • $\begingroup$ I must admit I'm struggling really badly with this. The only thing I can think is if $\delta = \frac{1}{\epsilon}$ but I don't think this is correct... $\endgroup$
    – Killuminati
    Mar 15, 2016 at 18:26
  • $\begingroup$ Just try to solve Hint 1... $\endgroup$
    – J.-E. Pin
    Mar 15, 2016 at 18:28
  • $\begingroup$ I am completely lost. I have tussled with this all night... Scrambling through notes and the joys of the internet I'm still no closer to what the answer is. Any other $\varepsilon - \delta$ continuity question I've solved I have done so going backwards from $|f(x) - f(c)| < \varepsilon$ but I am continually hit by $0 < \varepsilon$ using this method. And I can't seem to work out how to choose \delta as in hint 1. My guesses just don't make sense i.e $\delta = \frac{1}{c}$ etc. I'm going to have to admit defeat, thank you for your help anyway! $\endgroup$
    – Killuminati
    Mar 16, 2016 at 12:55
  • $\begingroup$ Try $\delta = \min(|c|/2, (1 - |c|)/2)$. $\endgroup$
    – J.-E. Pin
    Mar 16, 2016 at 14:04

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