A trivial question about continuity

Question

$f:\mathbb{R}\longrightarrow\mathbb{R}:x\mapsto x^2$

This function is continuous as we all know.

Since for every point in the domain, we will always be able to draw a $\delta\epsilon-$rectangle, for every $\epsilon$ which captures every point of $f(x)$ if it captures $x$.

As i first started looking on continuouity I thought it exists to help clearify weither a curve has "holes". Or points where the curve diverges. It made sense to me, but then I thought what does it mean to be pointwise continuous?

$h:\lbrace 1,2,3 \rbrace\longrightarrow\mathbb{R}:x\mapsto x^2$ is this function also continuous? Why shouldnt it be? I mean if i can pointwise look at a curve and take all the points out of the domain where it is uncontinuous, the curve will be continuous. But I could also take alot of points where the function is continuous out of the domain and it should stay continuous. Even if I leave the function only with a set of ,for example, 3 elements in the domain. Is this correct?

And if yes, what is the advantage of stating a function like $h$ is still continuous, if it has nothing left to do with curves.

So why do we need this property? Shouldnt it also has some value for $h$ to be continuous?

I know this question is really basic, and stupid, but it really interests me to understand what "pointwise-continuity really means for a function. Thank you

Answer 1

The standard definition of continuity has the following convenient property:

If $f \colon X \to Y$ is continuous and $A$ is a subset of $X$ then the restriction of $f$ to $A$ is continuous.

This remains true no matter how bizarre $A$ is - $X$ could be $\mathbb{R}$ and $A$ could be the Cantor set, or a non-measurable set, or a finite set as in your example.

So I suggest tweaking your intuition about continuity. The normal intuition is something like "I can draw the graph without lifting my pencil", but this intuition falls apart if you can't even draw the domain of the function without lifting your pencil! Instead, try to think of continuity as something like "$f$ is continuous if it respects the local structure of its domain" If the domain doesn't have much local structure, e.g. if it's a discrete set, then this isn't saying very much - but that's the domain's fault, not $f$'s.

Answer 2

You're right that continuity is a property of each point of the domain. Usually when we say a function $f$ is continuous, we just want to say it is continuous on all its points.

In the $\epsilon-\delta$ definition of continuity you mentioned, the domain and range of a function are taken into account. A function $f : \{1,2,3\} \rightarrow \mathbb{R}$ will always be continuous under that definition. For instance, to show its continuity in $f(1)$, just take $\delta = \frac{1}{2}$ for any $\epsilon > 0$ and all points within distance $\frac{1}{2}$ of 1 in the domain (which is just 1) will fall within distance $\epsilon$ in the range.

The example above shows that yes, you can remove enough parts of the domain to make a function continuous. This does not necessarily imply the original function is continuous on the bigger domain.

Answer 3

So, we need to be careful. Continuity is first and foremost a topological term. In the category of topological spaces, continuous maps are the morphisms and topological spaces the corresponding objects.

That means, in order to be able to talk about continuity, we need to make sure the spaces we are talking about are in fact topological spaces.

The definition of continuity in the language of topological spaces is the following:

Let $X,Y$ be topological spaces with topologies $\mathcal{T}_X$ and $\mathcal{T}_Y$ respectively. A map $f:X\to Y$ is continuous, if for every open set $V\subset Y$ in $\mathcal{T}_Y$, the set $f^{-1}(V) \subset X$ is open in $\mathcal{T}_X$.

So you notice, continuity does not "come out of thin air", but is dependend on the given topological structure the corresponding sets admit.

Given your example of $\{1,2,3\}$ we can actually define a topology $\mathcal{T}$ on it by declaring its open sets to be $\{1,2,3\}, \varnothing, \{1\}, \{2\},\{3\}$. and any union of these elements. To sum it up we get

$$\mathcal{T} = \{\{1,2,3\}, \varnothing, \{1\}, \{2\},\{3\},\{1,2\},\{1,3\},\{2,3\}\}$$

Now we can actually use this topology $\mathcal{T}$ on $\{1,2,3\}$ together with your map $f:\{1,2,3\}\to \mathbb{R}$ to induce a topology on $\mathbb{R}$ (called the final topology) via

$$\mathcal{T}_\mathbb{R} = \{ U \subset \mathbb{R} \mid f^{-1}(U) \in \mathcal{T}\}$$

What we are doing here is, we define the open subsets of $\mathbb{R}$ exactly to be those subsets, for which $f^{-1}(U)$ is an element of $\mathcal{T}$, i.e. of the given topology on $\{1,2,3\}$.

Now how do the open sets of $\mathbb{R}$ with respect to $\mathcal{T}_\mathbb{R}$ look like?

Considering that we have $$f\colon \{1,2,3\} \mapsto \mathbb{R},\ x \mapsto x^2$$

we get $$\mathcal{T}_\mathbb{R} =\{ \mathbb{R}, \varnothing,\{1\},\{4\},\{9\},\{1,4\},\{1,9\},\{4,9\},\{1,4,9\} \}$$

Now by construction, your map $f$ is actually continuous on $\{1,2,3\}$! That's simply because we build the topology on $\mathbb{R}$ in exactly the way that the preimage of $f$ for every declared open subset of $\mathbb{R}$ is an open subset of $\{1,2,3\}$.

Now you can't possibly draw your function $f$ in the same way you would expect it to look like as if we had a function $f:\mathbb{R} \to \mathbb{R}$.

But any visualization is fine, as long as it reflects what's actually happening, namely the preimage of every open subset in $\mathcal{T_\mathbb{R}}$ is an element of $\mathcal{T}$.

Do not hesitate to ask questions, if you have any.