Definition of a Continuous Functions

[Oxymoron]

There are several definitions of a continuous function between metric spaces. Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and let $f:X\rightarrow Y$ be a function. Then we have the following as definitions for continuity of $f$:

$$\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } d_X(y,x) < \delta \, \Rightarrow \, d_Y(f(y),f(x)) < \epsilon$$

$$\square \quad \forall\, x \in X \mbox{ and } \forall \, \epsilon >0 \, \exists \, \delta > 0 \mbox{ such that } f(B(x,\delta)) \subseteq B(f(x),\epsilon)$$

$$\square \quad \mbox{For every open subset } U \mbox{ of } Y \mbox{ the inverse image } f^{-1}(U) \mbox{ is an open subset of } X$$

I want to prove that each of these statements are equivalent to each other. When I do this, is it okay to prove (1) <=> (2), (2) <=> (3), and (3) <=> (1), that is, I don't have to prove (2) => (1) once I have proven (1) => (2).

Secondly, is it okay if I just prove that each statement proves that $f$ is continuous and hence they are equivalent (ie. the statements are equivalent because they each prove continuity of $f$?). If this is not the right way to prove that each statement is equivalent, could someone tell me how I would do that.

 

[Galileo]

Whenever you want to prove that a list of statements are equivalent, like here, the most efficient way is usually proving that (1) => (2), (2) => (3) and (3) => (1).
The drawback is however that, assuming (1), you can only understand why (3) is true via (2).

So proving (1)<=> (2), (2)<=>(3) and (3)<=>(1) is okay, but are 6 proofs where you logically only need 3. (I do advice to try it though, it's good exercise).

When I do this, is it okay to prove (1) <=> (2), (2) <=> (3), and (3) <=> (1), that is, I don't have to prove (2) => (1) once I have proven (1) => (2).

I`m not sure what you mean. You don't have to prove (2)=>(1) if you have proven (2)=>(3) and (3)=>(1), but you can't assume (2)=>(1) is true from just proving (1)=>(2).

Secondly, is it okay if I just prove that each statement proves that is continuous and hence they are equivalent (ie. the statements are equivalent because they each prove continuity of ?). If this is not the right way to prove that each statement is equivalent, could someone tell me how I would do that.

No, this is not okay. You have to start with a definition of a continuous function if you want to prove that a function is continuous. So you have to start with one and prove that it is equivalent to another. Also, if you start with the definition from (1) and then prove that (2) and (3) implies the functions are continuous, you've proven (2)=>(1) and (3)=>(1), not the other ways around.

 

[Oxymoron]

Thankyou Galileo for replying.

I made quite a few mistakes in my first post.

I had a go a trying to prove (2) => (3).

Let $U$ be open in $Y$. Therefore there exists an $\epsilon > 0$ such that $B(f(x),\epsilon) \subset U$. From (2) we know that $\exists \, \delta > 0$ such that $f(B(x,\delta)) \subset B(f(x),\epsilon)$. Therefore

$$B(x,\delta) \subset f^{-1}(f(B(x,\delta))) \subset f^{-1}(U)$$

Therefore $f^{-1}(U)$ is open in $X$. Which imples (3).


How does this sound?

 

[Galileo]

Well on your way :)

The best thing is scrutinize the proof for any details that might be considered unclear.
For example, you haven't stated what f(x) is in your proof. You just said: Let U be an open set in Y, then there exists such and such that B(f(x),e) is a subset of U.
The point is ofcourse that there for every point $a\in U$ exists an $\epsilon$ such that $B(a,\epsilon) \subset U$ (by the definition of open set). This raises the question: Can every point in U be written as f(x) for some x in the domain of f?
Other than that, it looks good.

 

[Oxymoron]

$f(x)$ is the image point in $Y$ of the point $x \in X$.


I have written down exactly what was going through my head...

The point is ofcourse that there for every point exists an such that (by the definition of open set). This raises the question: Can every point in $U$ be written as $f(x)$ for some $x$ in the domain of $f$?

The way I see it is that $U$ is an open subset in $Y$, ie $U \subset Y$ is open. Because of this, I can construct an open ball with radius $\epsilon$ around any image point $f(x) \in U$, such that the open ball is contained within $U$. Formally speaking, there exists a positive, non-zero $\epsilon$ such that $B(f(x),\epsilon) \subset U$.

But from (2) we can look backwards from the image point, ie. look at where the inverse mapping takes the image point $f(x)$. In fact, we know that we can construct an open ball around $f^{-1}(f(x))$ with positive, non-zero radius $\delta$ such that

$$f(B(x,\delta)) \subset B(f(x),\epsilon)$$

In other words, the function maps the open ball $B(f(x),\epsilon) \subset U \subseteq Y$ into an open ball $f^{-1}(B(f(x), \epsilon))$.

So we have two open balls; one in $X$ and one in $Y$. We know that they are both open, one is centered at $x$ and the other at $f(x)$. The function $f^{-1}$ maps open balls in $Y$ to open balls in $X$ (because of (2)). Now comes the crucial point:

Say we construct an open ball, $B(x,\delta)$, around $x \in X$ with radius $\delta$. Then apply the function. The image of this ball will be contained within the ball around $f(x)$ with radius $\epsilon$! Since I can always change my value for $\epsilon$ to compensate for any $\delta$. Therefore

$$f(B(x,\delta)) \subset B(f(x),\epsilon)$$

Furthermore, if we map our open ball $B(x,\delta)$ into $Y$ and then back into $X$ (ie. we have $f^{-1}(f(B(x,\delta)))$ then the open ball $B(x, \delta)$ will always be contained within the image of the image, ie

$$B(x,\delta) \subset f^{-1}(f(B(x,\delta)))$$

But we know that $f(B(x,\delta)) \subset U$. Hence

$$B(x,\delta) \subset f^{-1}(U)$$

Therefore for every point $a \in f^{-1}(U)$ we can construct an open ball $B(a,\delta)$. Therefore by definition, $f^{-1}(U)$ is open in $X$.


Does this sound better?

 

[Oxymoron]

Can every point in $U$ be written as $f(x)$ for some $x$ in the domain of $f$?

Yes. This comes from (3) => (1)

Suppose $f^{-1}(U)$ is open in $X$.

Let $x \in X$ and $\epsilon > 0$. Since $B(f(x),\epsilon)$ is open in $Y$, and the set $f^{-1}(B(f(x),\epsilon))$ is open in $X$.

Since $x \in f^{-1}(B(f(x),\epsilon))$, there exists a $\delta > 0$ such that $B(x,\delta) \subset f^{-1}(B(f(x),\epsilon))$.

This implies that $f(B(x,\delta)) \subset B(f(x),\epsilon)$.

But this means that $d(x,x_0) < \delta \Rightarrow d(f(x),f(x_0)) < \epsilon$ for all $x,x_0 \in X$.

Since $B(x,\delta) \equiv d(x,x_0) < \delta$ for all $x \in X$ and some $x_0 \in X$.

Similarly $B(f(x),\epsilon) \equiv d(f(x),f(x_0)) < \epsilon$.