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Let $A\subset \mathbb{R}^n$ be a set, $f:A\to\mathbb{R}^m$ be a function, $a\in A$. Suppose that $f$ is sequentially continuous at $a$. Prove that $f$ is continuous at $a$ in the sense of $\varepsilon-\delta$ definition.

My proof:

Let $\{x_k\}\subset A$ be a sequence such that $x_k$ converges to $a$. Then, since $f$ is sequentially continuous on $A$, $\forall \varepsilon >0$, $\exists k_0\in\mathbb{N}$ such that $\|f(x_k)-f(a)\|<\varepsilon$ whenever $k\ge k_0$. Also, $\forall\delta >0$, $\exists k_1\in\mathbb{N}$ such that $\|x_k-a\|<\delta$ whenever $k\ge k_1$. Let $k':=\max\{k_0, k_1\}$, let $k\ge k'$ (we can fix $k$), then $\|x_k-a\|<\delta\implies \|f(x_k)-f(a)\|<\varepsilon$. Hence, $f$ is continuous at $a$ in the $\varepsilon-\delta$ sense.

The marker deducted about 67% of the marks and commented that my argument is done for one sequence convergent to $a$ instead of covering all sequences convergent to $a$. However, I don't understand this argument because $x_k$ was taken to be an arbitrary sequence converging to $a$. So what is my mistake then?

sequence
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    Your argument shows that and only that there are countably many "working" points close to $a$ by $\delta$ instead of that every point close to $a$ by $\delta$ works. – Yes Jun 28 '17 at 18:59
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    @YngwieMalmsteen Can you please clarify what you mean? – sequence Jun 28 '17 at 19:07
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    I can't understand the marker’s comment either; anyway, you failed to prove the statement. Nothing in your proof can be salvaged, I'm afraid. – egreg Jun 28 '17 at 20:06

1 Answers1

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Your mistake was to pick a sequence, then to pick $\varepsilon > 0$, then to pick $k_{0}$, then to pick $\delta > 0$ "satisfying the definition of continuity for your sequence".

Instead you need to show "there exists a single $\delta > 0$ that 'works' for every sequence". The natural approach here is contrapositive: Assume $f$ is not $\varepsilon$-$\delta$ continuous at some point $a$, and to construct a sequence $(x_{k}) \to a$ such that $(f(x_{k})) \not\to f(a)$, so that $f$ is not sequentially continuous at $a$.

Andrew D. Hwang
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  • But can't we take $\delta>0$ as small as we wish? Then such a $\delta$ would fit every sequence, for $k$ greater than some large $k_0$. – sequence Jun 28 '17 at 19:45
  • In your proof, you've picked $\delta > 0$ after you've picked your sequence. There are infinitely many sequences (to put it mildly), and the choice of $\delta$ is only "allowed" to depend on ($\varepsilon$ and) $a$ and the function $f$, not on a particular sequence. – Andrew D. Hwang Jun 28 '17 at 20:12
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    Sorry for my misunderstanding, I really want to clarify this matter once and forever. I think there is a subtle peculiarity I'm not yet conceiving of. I still don't completely understand my principal error. Can you please clarify why my approach is not consistent with the $\varepsilon-\delta$-definition? The definition says: $f$ is continuous at $a\in A$ if for every $\varepsilon>0$ we can find a $\delta > 0$ such that, for all $x\in A$, $|x-a|<\delta$ implies $|f(x)-f(a)|<\varepsilon$. So what I did was to take a $\delta>0$ such that $|x-a|<\delta$ and also $|f(x)-f(a)|<\varepsilon$. – sequence Jun 28 '17 at 22:16
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    Also, can you please clarify what you mean by "to put it mildly"? Can there be more than infinitely many sequences? – sequence Jun 28 '17 at 22:17
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  • What you've shown is, "If $(x_{k}) \to a$, then for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|x_{k} - a| < \delta$, then $|f(x_{k}) - f(a)| < \varepsilon$". What you want is, "If $|x - a| < \delta$, then $|f(x) - f(a)| < \varepsilon$". The problem is, not every element of $A$ in the $\delta$-ball centered at $a$ is a term of your sequence. (That may help explain the marker's comment.) 2. By "mildly", I meant that if $A$ is an uncountable set, the image of a sequence is never an open subset of $A$, hence never the intersection of $A$ with an open ball. :)
    • – Andrew D. Hwang Jun 29 '17 at 00:01
  • Thanks, now it's clearer :) However, I'm still wondering that if $x_k$ are in the $\delta$-ball, then don't there exist infinitely many sequences converging to $a$ so as to cover the entire $\delta$-ball? I realize, however, that there will generally be a different $\delta$ for different sequences; and since there are infinitely many sequences, we cannot take a maximum delta (but can there exist something like the supremum of such deltas?). It looks like this thing is quite peculiar, but very useful to know for the future. – sequence Jun 29 '17 at 06:17
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    If you "take the $\delta$s corresponding to two sequences", their minimum "works for both sequences". As you say, you can pick (uncountably many) sequences to cover a ball about $a$; the potential problem is, the infimum of the resulting $\delta$s might be $0$. Your task, if you like, is to show this is no obstacle: You can "pick a single $\delta > 0$ that 'works' for every sequence". My answer to When to use the contrapositive to prove a statment may help explain strategically why a contrapositive argument is "more natural" here. – Andrew D. Hwang Jun 29 '17 at 09:51
  • Can you please clarify why the minimum $\delta$ would have to be considered and not the maximum one? For if we have two deltas, $\delta_1, \delta_2$, say, with $\delta_1<\delta_2$, then if $|x-a|<\delta_1$ then $|x-a|<\delta_2$, so that $\delta_2$ necessarily satisfies the condition, while, if vice versa, then $\delta_1$ would not necessarily satisfy the condition. – sequence Jun 29 '17 at 21:39
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    The definition of continuity contains the clause "If $|x - a| < \delta$, then $|f(x) - f(a)| < \varepsilon$". If you know that the conclusion holds provided $|x - a| < \delta_{1}$ (for some collection of $x$s) and holds for $|x - a| < \delta_{2}$ (for some other collection of $x$s), and you want a single $\delta$ to encompass both cases, you need to ensure $\delta \leq \delta_{i}$ for $i = 1, 2$, i.e., $\delta \leq \min(\delta_{1}, \delta_{2})$. – Andrew D. Hwang Jun 29 '17 at 21:54
  • But if something holds for $\delta_1$ then it also holds for $\max(\delta_1, \delta_2)$, correct? – sequence Jun 29 '17 at 22:23
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    In this context, "something holds for $\delta_{1}$..." is too non-specific. :) What you want is analogous to "If $|x - a| < \delta$, then $|x - a| < \delta_{1}$ and $|x - a| < \delta_{2}$, so...". For that, $\delta$ needs to be at most the smaller of the two. – Andrew D. Hwang Jun 29 '17 at 22:44
  • I see now. But then it is the question of how a $\delta$ is picked for every $\varepsilon$. That is, it appears to be that a delta needs to be of some size (depending on the $\varepsilon$), and that size should not be greater than some limit. Correct? Otherwise, we would be able to pick a maximum $\delta$ and not worry about the rest of the deltas. – sequence Jun 30 '17 at 01:47
  • Let us continue this discussion in chat. – Andrew D. Hwang Jun 30 '17 at 10:45