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I am looking for proof of $+$ and $\times$ are continuous operations without using the standard definition of continuity (1. $\epsilon-\delta$, or 2. preimage of open sets or 3. sequential continuity) Any other proof would be considered valid for this purpose

Define: Addition and multiplication as mappings

$+: \mathbb{R} \times \mathbb{R} \to \mathbb{R}, (a,b) \mapsto a+b$

$\times: \mathbb{R} \times \mathbb{R} \to \mathbb{R}, (a,b) \mapsto a \times b$

Using obscure definition of continuity:

$f$ is continuous if $\forall A \subset \mathbb{R} \times \mathbb{R}, f(\overline A) \subseteq \overline {f(A)}$

We know that singletons are closed in $\mathbb{R}$, closed sets are closed under cartesian product, so $A = \{a\} \times \{b\} \subset \mathbb{R} \times \mathbb{R}$ is closed

$f(\{a\} \times \{b\}) = f(\overline{\{a\} \times \{b\}}) = \overline{f(\{a\} \times \{b\})}$

The latter equality because $f(\{a\} \times \{b\})$ is a singleton, so $f$ is continuous

Can someone please check on the validity? And perhaps offer an alternative proof, thanks

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Fraïssé
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    Are you asking whether your proof works (which it doesn't, because it only handles the case where $A$ consists of $1$ point - and applies to all functions $f$, regardless of continuity), or for other proofs of continuity? – Milo Brandt Jun 04 '16 at 21:49
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    That definition of continuity (which is hardly obscure) involves an arbitrary set $A$, not just singletons. – anomaly Jun 04 '16 at 22:11
  • Why not just use sequential continuity and the rules for limits of sequences: https://en.wikipedia.org/wiki/Limit_of_a_sequence – Maik Pickl Jun 04 '16 at 22:13
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    As pointed out above you have not met the requirements of the "obscure definition" of continuity. For separable metric spaces, that definition is just sequential continuity in an elaborate disguise. You need to say more about your motivation for asking the question. – Rob Arthan Jun 04 '16 at 22:14
  • The whole point is to prove this without the usual definitions.. – Mathemagician1234 Jun 04 '16 at 22:14

1 Answers1

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Here's an easy way to tell that something must be wrong with your proof: what sort of assumptions on $f$ does it use? Your proof doesn't assume anything about $f$, so - if it were valid - would prove that every function is continuous.

The flaw, as the comments have pointed out, is that you need to show $$f(\overline{A})\subseteq\overline{f(A)}$$ for every $A\subseteq \mathbb{R}^2$, not just those of the form $\{a\}\times\{b\}$. Here's an example which might make this easier to visualize (on $\mathbb{R}$, instead of $\mathbb{R}^2$, for simplicity):

  • Let $f(x)$ be the characteristic function of $\mathbb{Q}$: $1$ if $x$ is rational, $0$ if $x$ is not rational. (This is called the Dirichlet function.)

  • Take $A=\mathbb{Q}$. What is $\overline{A}$? What is $f(\overline{A})$?

  • On the other hand, what is $f(A)$? What is $\overline{f(A)}$?

In terms of using this definition to prove things, I think it's generally faster to first prove that this is equivalent to the $\epsilon$-$\delta$ definition, and then use that one; in most cases that makes things a little more concrete and easy to figure out (at least early on). Of course, Your Mileage May Vary.

Noah Schweber
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  • Thanks, I didn't pose this question, someone posed this question to me and claimed that a short proof can be found not via the standard definitions but through their consequences and equivalent definitions.. – Fraïssé Jun 04 '16 at 22:28