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The real function $f(x)$ defined so that $f(x)=x$ when $x \in \mathbb{Q}$ and $f(x)=1 $ when $x \notin \mathbb{Q}$ is Weierstrass continuous.

But, it doesn't have sequential continuity. If you try the sequence $(x_n)$ of rationals that converges to $\pi$ or any real number $c$ for that matter, you realize that there's no $N$ that's large enough so that $\forall \epsilon > 0, |f(x_N)-c|<\epsilon$ where $c$ is considered the limit of that particular sequence.

So, it appears that Weierstrass continuity doesn't imply sequential continuity.

  • Yes, it's the $\epsilon-\delta$ definition of continuity. Why do you say it's wrong? $\mathbb{Q}$ is dense in $\mathbb{R}$ –  Oct 10 '15 at 19:37
  • What's the definition of Weierstrass continuous you are using? If it's the classical $\epsilon$-$\delta$-continuity then the assertion does not hold. – user159517 Oct 10 '15 at 19:37

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Your function isn't continuous (except at $1$).

Weierstrass continuity is the same thing as sequential continuity in first-countable spaces (thus in metric spaces in particular, and more specifically, in $\mathbb R$ with the usual metric).

However, in general topological spaces, the two are not the same thing: a W-continuous function is necessarily sequentially continuous, but the converse isn't necessarily true (and that's why in general topological spaces we use the notions of nets instead of sequences).

Anyway, the final answer to your question is: W-continuity always implies sequential continuity, but not the other way around.

  • Ok. I thought about it more, tried epsilon equals 1, and I realize that I was wrong... –  Oct 10 '15 at 20:06