Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider this definitions:

A function $f:X \to Y$ is continuous at $x\in X$ iff for any open neighborhood $V_{f(x)}$ of $f(x)$ there is an open neighborhood $U_{x}$ of $x$ that gets mapped by $f$ into $V_{f(x)}$ (or, in other words, there is an open neighborhood $U_{x}$ of $x$ such that $f[U_x]\subseteq V_{f(x)}$). A function $f:X \to Y$ is continuous iff it is continuous at all $x \in X$.

This definition of continuity seems to me equivalent to the "standard" definition in terms of inverse images (namely $f:X\to Y\;$ is continuous iff for any open set $V\subseteq Y$, the set $f^{-1}(V)\subseteq X$ is open).

Am I wrong?

Assuming that I'm correct, I am baffled by the prevalence of the currently standard definition, since the one above looks to me far more natural. It is certainly more obviously a generalization of the $\epsilon$-$\delta$ definition of continuity in metric spaces (just replace $V_{f(x)}$ and $U_{x}$ by open balls $B(f(x), \epsilon)$ and $B(x, \delta)$, respectively), which, in turn, is an obvious generalization of the $\epsilon$-$\delta$ definition of continuity for functions $\mathbb{R}\to\mathbb{R}$ (just replace the open balls $B(f(x), \epsilon)$ and $B(x, \delta)$ by open intervals $(f(x) - \epsilon, f(x) + \epsilon)$ and $(x-\delta, x+\delta)$, respectively).

Given these considerations, why is the standard definition the generally preferred one?

Edit: the replies I've gotten so far have focused on the fact that the alternative definition depends on an auxiliary definition of continuity at a point, but this is a very minor aspect of the alternative definition. I chose this two-part approach only to make the wording of the definition of continuity slightly less awkward, but it is not required. I could have just as well written:

A function $f:X \to Y$ is continuous iff for all $x \in X$ and any open neighborhood $V_{f(x)}$ of $f(x)$ there is an open neighborhood $U_{x}$ of $x$ such that $f[U_{x}]\subseteq V_{f(x)}$.

Also, these replies suggest that, when it comes to defining terms, brevity trumps clarity. I find this hard to take: a definition, by definition, is introducing a concept, so its intended audience is one that will appreciate clarity over brevity. An equivalent characterization of the same concept whose only advantage is greater brevity should be relegated to a theorem, IMO.

share|improve this question
    
I hope I'm reading it correctly, though I think you're just saying that given e > 0 there exists a d > 0 s.t. f(B(d,x)) is a subset of B(e,f(x)) ? Right? The preimage definition has tremendous importance as it generalizes the concept of nhoods to that of cts functions preserving open sets. –  Adam Jan 4 '12 at 12:29
2  
A mathematician, writing an argument that requires a proof of continuity, will sometimes use that characterization. And (except at the student level) there would be no special comment that this is equivalent to some other definition. –  GEdgar Jan 4 '12 at 14:10
2  
Dear kjo: A topology can be defined by specifying all the open sets, or the neighborhoods of each point, or the closure of each subset. I'd expect your definition to be in Bourbaki. Indeed, the notion of filter is one of their specialities. I don't see what's nonstandard in your definition. Of course, your point is an outstanding one! –  Pierre-Yves Gaillard Jan 4 '12 at 15:29
    
You make an interesting point about filters, @Pierre-Yves. Can you elaborate the connection to them in an answer? –  Srivatsan Jan 4 '12 at 15:54

3 Answers 3

Your definition is perfectly fine. It is mentioned (as a theorem) in Munkres, for instance.

The standard definition is simpler. Moreover, in topology there is less emphasis on continuity at a point than in analysis.

share|improve this answer
    
For the record, Munkres may mention the alternative definition somewhere in his textbook, but the "official" definition he gives (p. 102 of 2nd edition) is the standard one: "A function $f:X\to Y\;\;$ is said to be continuous if for each open set $V$ of $Y$, the set $f^{-1}(V\;)$ is an open subset of $X$." –  kjo Jan 4 '12 at 14:48
    
@kjo, thanks for clarification. –  lhf Jan 4 '12 at 15:18

Apart from the reasons already mentioned, I like the pre-image definition because once you get used to it, it is actually simpler and clearer than your definition. One case in point is that it is almost completely free of quantifiers, as compared to

For all $x$ and for all $V_{f(x)}$, there exists an open neighborhood $U_x$ such that...

On a general note, an excessive use of quantifiers has multiple disadvantages:

  • Inexperienced students often get confused in manipulating quantifiers, which results in great many buggy proofs. In fact, one can find a number of questions in MSE testifying to this.

  • Quantifiers tend to obscure, rather than clarify, the intuitive meaning of the concept being defined. This is because we are not wired up to think or reason about three levels of quantifiers.

  • Even after I got comfortable with the definition, I found it quite arbitrary. Why is this the “correct” order for the quantifiers? What happens if you switch the $\exists$ and $\forall$ quantifiers? Most -- but not all -- of such switches lead to either trivial or strange definitions and should not be considered seriously; but that raises the question how one could recognise the right order.

I would even venture to say that if your definition involves too many layers of quantifiers, then you are likely not doing it “right”. It's time to revisit your definition! :-)

share|improve this answer

The standard definition is standard because one usually only needs to be able to talk about global continuity of functions rather than continuity at each point. Given that this is the case, the standard definition makes good sense since it's the cleanest possible: it makes no references to anything other than the topologies of the involved spaces.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.