I can say “sin(1/x) is a continuous function on (0,1] but its graph is not path connected”, which is more formal, but likely not mean anything to most of the reader. In that sense, I guess I have also lied :)

It’s also false. Take any pair of points on the graph of sin(1/x) using the domain (0,1] that you just gave. Then we can write these points in the form (a,sin(1/a)), (b,sin(1/b)) such that 0 < a < b without loss of generality. The map f(t)=(t,sin(1/t)) on [a,b] is a path connecting these two points. This shows the graph of sin(1/x) on (0,1] is path connected.

This same trick will work if you apply it to the graph of ANY continuous map from a connected subset of R into R. This is what my graph example was getting at.

The “topologists sine curve” example you see in pointset topology as an example of connected but not path connected space involves taking the graph you just gave and including points from its closure as well.

Think about the closure of your sin(1/x) graph here. As you travel towards the origin along the topologists sine curve graph you get arbitrarily close to each point along the y-axis between -1 and 1 infinitely often. Why? Take a horizontal line thru any such point and look at the intersections between your horizontal line and your y=sin(1/x) curve. You can make a limit point argument from this fact that the closure of sin(1/x)'s graph is the graph of sin(1/x) unioned with the portion of the y-axis from -1 to 1 (inclusive).

Path connectedness fails because there is no path from any one of the closure points you just added to the rest of the curve (for example between the origin and the far right endpoint of the curve).

A better explanation of the details here would be in the connectedness/compactness chapter in Munkres Topology textbook it is example 7 in ch 3 sec 24 pg 157 in my copy.

However, I like to push back on the assumption that, in the context of teaching continuous function, the underlying space needs to be bounded: one of the first continuous function student would encounter is the identity function on real, which has both a infinite domain and range.

This is fine. I stated boundedness as an additional assumption one might require for pragmatic reasons. It’s not mandatory. But it’s easy to imagine somebody trying to be clever and pointing out that if we allow the domain or range to be unbounded we still have problems. For example you literally cannot draw the identity function in full. The identity map extends infinitely along y=x in both directions. You don’t have the paper, drawing utensils or lifespan required to actually draw this.

More impressively, you can have function that is continuous, but you cannot find a connected path on it (i.e. not path connected). In plain words, if anyone told you “a function is continuous when you can draw it without lifting your pen”. They have lied to you.

You are misrepresenting an analogy as a lie. Besides that, in the context where the claim is typically made, the analogy is still pretty reasonable and your example is just plain wrong.

People are talking about continuous maps on subsets of R into R with this analogy basically always (i.e., during a typical calc 1 or precalc class). The only real issue are domain requirements in such a context. You need connectedness in the domain or else you’re just always forced into lifting your pen.

There are a couple other requirements you could add as well. You might also want to avoid unbounded domains since you can’t physically draw an infinitely long curve. Likewise you might want to avoid open endpoints or else things like 1/x on (0,1] become a similar kind of problem. But this is all trivial to avoid by saying “on a closed and bounded interval” and the analogy is still fairly reasonable without them so long as you keep the connectedness requirement.

For why your example is just wrong in such a context, say we’re only dealing with continuous maps on a connected subset of R into R. Recall the connected sets in R are just intervals. Recall the graph of a function f with domain X is the set {(x,f(x)) : x is in X}. Do you see why the graph of such a function is always path connected? Hint: Pick any pair of points on this graph. Do you see what path connects those two points?

Once you want to talk about continuous maps between more general topological spaces, things become more complicated. But that is not within the context in which this analogy is made.

There are some subtleties to this particular topic that are worth mentioning. I would be careful to distinguish between constructing vs defining here.

The usual definition of the irrationals works roughly like this:

You have a set of numbers R which you call the real numbers. You have a subset of the real numbers Q which you call the rational numbers. You define a real number to be irrational if it is not a rational number.

This is perfectly rigorous, but it relies on knowing what you mean by R and Q.

Both R and Q can be defined “without” (a full) construction by letting R be any complete ordered field. Such a field has a multiplicative identity 1 by definition. So, take 0 along with all sums of the form 1, 1+1, 1+1+1 and so on. We can call this set N. We can take Z to be the set of all elements of N and all additive inverses of elements of N. Finally take Q to be the set containing all elements of Z and all multiplicative inverses of (nonzero) elements of Z. Now we have our R and Q. Also, each step of the above follows from our field axioms. Defining irrationals is straightforward from this.

So, the definition bit here is not a problem. The bigger issue is that this definition doesn’t tell us that a complete ordered field exists. We can define things that don’t exist, like purple flying pigs and so on.

What the dedekind cut construction shows is that using only the axioms of zfc we can construct at least one complete ordered field.

So its a case of it not working on irrational numbers, its just that we cant prove it because we cant calculate the multiplication of 2, right?

The issue is the proving part. We can’t use repeated addition trickery (at least not in an obvious way) to show a product of two irrational negative numbers is positive. It’s definitely still true that a product of two negative numbers is positive, just that proving it in general requires a different approach.

Somehow, my mind has issues with the e*pi = ke. Id say that ke = e * pi is impossible because k is an integer and pi isnt, no? It could never be equals, i think.

Yes this is correct. The ke example is for a proof by contradiction. We are assuming something is true in order to show it forces us to be able to conclude something ridiculous/false. Since the rest of our reasoning was correct, then it must have been our starting assumption that was wrong. So, we have to conclude our starting assumption was wrong/false.

Multiplying two negative irrational numbers together will still give you a positive number, it’s just that you can’t prove this by treating multiplication as repeated addition like you can multiplication involving integers (note that 3 is an integer, 3 is not irrational, the issue is when you have two irrationals).

So, for example with e * pi, pi isn’t an integer. No matter how many times we add e to itself we’ll never get e * pi.

Try it yourself: Assume that we can add e to itself k (a nonnegative integer) times to get the value e * pi. Then e * pi = ke follows by basic properties of algebra. If we divide both sides of this equation by e we find that pi=k. But we know k is an integer, and pi is not an integer. So, we have reached a contradiction and this means our original assumption must be false. e * pi can’t be equal to e added to itself k times (no matter which nonnegative integer k that we pick).

For reference:

https://en.m.wikipedia.org/wiki/Real_numbers

https://en.m.wikipedia.org/wiki/Ordered_field

Copy pasted from my other comment:

This doesn’t work if you have to deal with multiplication of numbers that are not integers. You can adjust your idea to work with rational numbers (i.e. ratios of integers) but you will have trouble once you start wanting to multiply irrational numbers like e and pi where you cannot treat multiplication easily as repeated addition.

The actual answer here is that the set of real numbers form a structure called an ordered field and that the nice properties we are familiar with from algebra (for ex that a product of two negatives is positive) can be proved from properties of ordered fields.

For reference:

https://en.m.wikipedia.org/wiki/Real_numbers

https://en.m.wikipedia.org/wiki/Ordered_field

This doesn’t work if you have to deal with multiplication of numbers that are not integers. You can adjust your idea to work with rational numbers (i.e. ratios of integers) but you will have trouble once you start wanting to multiply irrational numbers like e and pi where you cannot treat multiplication easily as repeated addition.

The actual answer here is that the set of real numbers form a structure called an ordered field and that the nice properties we are familiar with from algebra (for ex that a product of two negatives is positive) can be proved from properties of ordered fields.

Don’t confuse the wording “set of real numbers” here, this is just the technical name for the collection of numbers people use from elementary algebra on through to calculus.

People talk a lot about stackoverflow for figuring out bugs and miscellaneous coding questions but the whole stackexchange project has a lot of other very excellent websites.

Helping “bad people”? There’s nothing inherently more bad about Palestinians than any other nationality.