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continuous at
continuous function, continuous map, continuous mapping
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26A15 no label found54C05 no label found81-00 no label found82-00 no label found83-00 no label found46L05 no label found


would it be possible to have a more elementary definition that doesn't involve metric spaces ?

It is probably not possible to have a more elementary definition by omitting metric spaces. The most elementary possible definitions of continuous (that is, the ones you see in first year calculus texts) still need to use deltas and epsilons, and given that a correct definition needs to use deltas and epsilons, using the language of metric spaces adds little extra complexity.

The correct question, perhaps, is whether it is possible to give a more elementary definition not involving _topological_ spaces. I attempted to do just that in the second paragraph.

Some have the opinion that the nonstandard analysis definition of continuous (namely, a function is continuous if and only if infinitesimally close points in the domain have infinitesimally close images) is the simplest of all, but the nonstandard analysis approach has the drawback that you must define infinitesimals, and I think that no net effort is saved in the end by going this route.

I agree completely with you that the non-standard definition is not shorter or simpler once you take into account the fact that, to use it you first have to define non-standard numbers. To do that either reequires ultrafilers or model theory, and both of these are advanced topics.

However, I still thiink it would be a good idea to add the non-standard definition. My reason for saying this is that, if this is supposed to be an encyclopaedia, it should try to include everything, easy or hard. Maybe you could add a sentence like "In non-standard analysis, a function is defined to be continuous if ...". One of these days (if someone else doesn't do it first), I plan to add something about non-standard analysis. Probably, I'll lump it together with p-adic analysis in a topic entry on non-Archimedian analysis.

I think there is some sort of subtlety I'm missing, because it seems like the following example would be non-continuous under the current definition:

X = [0,1]
Y = Real numbers
f(x) = 0 (constant for all x in [0,1]).

Then if U = (-1,1), U is open and in Y, but yet F^-1(U) = {0} which is not open.

A constant function is not continuous!? What is wrong?

Whoops, means to say that F^-1(U) = [0,1]. Same problem still applies.

The set [0,1] is not open in the topology of R, but is open in the topology induced by (the Euclidean topology of) R on it.
And you moust think f as a function from the topological space [0,1] with the topology induced by (the Euclidean topology of) R, to the topological space R with the Euclidean topology.
Then f^-1((-1,1)) is the whole set [0,1]---which *is* open.

I'll try to give you a hint:

f must be a function between topological spaces. How is X=[0,1] a topological space? I mean, what topology were you thinking about when you picked it?

I'm sure you meant the topology inherited by the topology of Real numbers (search for "subspace topology", in case you are not familiar with it).

In this case, [0,1] is indeed an open set of X and your function is indeed continuous.. (Of course, this doesn't mean that [0,1] is an open set in R).

Think about it!

Oh wow, tricky how that works. Thanks all.

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