What is the intuitive meaning of continuity in topology?

[kade]

Quoted from Wikipedia,

A function

between two topological spaces X and Y is continuous if for every open set V ⊆ Y, the inverse image

is an open subset of X.
How to comprehend this definition in a intuitive way?

 

[olivermsun]

Kind of the same way you view continuity and intervals on the real line. A function $f$ is continuous at $x$ if for $y = f(x)$ you can stay as close as you want to $y$ by keeping the input close enough to $x$.

 

[pwsnafu]

Consider the statement "a continuous function preserves closeness". That means if $f(x)$ and $f(y)$ are close to each other then $x$ and $y$ were close to each other originally. If you want to study the function in a quantitative manner, that's metric spaces. Topology looks at the qualitative aspect.

 

[kade]

Kind of the same way you view continuity and intervals on the real line. A function $f$ is continuous at $x$ if for $y = f(x)$ you can stay as close as you want to $y$ by keeping the input close enough to $x$.

But, why its reverse image should be open in X?

Consider the statement "a continuous function preserves closeness". That means if $f(x)$ and $f(y)$ are close to each other then $x$ and $y$ were close to each other originally. If you want to study the function in a quantitative manner, that's metric spaces. Topology looks at the qualitative aspect.

Yep, I got it!

 

[kade]

Consider the statement "a continuous function preserves closeness". That means if $f(x)$ and $f(y)$ are close to each other then $x$ and $y$ were close to each other originally. If you want to study the function in a quantitative manner, that's metric spaces. Topology looks at the qualitative aspect.

But I think the continuous function doesn't have to be a bijection. So why should it, the inverse image of open set in Y, be open in X? Why not closed? Or anything else but disjointed.

 

[pwsnafu]

This is how I visualize it. (For now I'm assuming everything are connected open sets.)

Consider metric spaces. Let $U = (0,1)$ and consider two points, say, $x_0 = \frac{1}{12}$ and $x_n = \frac{9}{10}$. Because U is open, I can find an open neighborhood of $x_0$, namely, $B(1/12, 1/12)$. Pick a point to the right inside of this ball: for example $x_1 = \frac{1}{7}$. Now find a new open neighborhood: $B(1/7, 1/7)$ and find a new point $x_2 = \frac{3}{14}$ and repeat.

What we end up with is a sequence $x_0, x_1, x_2, \ldots, x_n$ where we "walk" from $x_0$ to $x_n$ such that

1. every point is in U
2. neighborhoods are open subsets of U
3. successive neighborhoods overlap with each other.

Now replace neighborhood with "open subset" and you can frame that procedure in terms of topology.
But this doesn't work with closed sets. Try $U = [0,1]$ and pick $x_0 = 0$. There are no open neighborhoods of $x_0$ which are also subsets of U. So we can't do our walking. This is a difference between open and closed with regards to this (very non mathematical and very arbitrary) concept of "closeness".

But what happens when $f^{-1}(V)$ is not connected. Let's suppose that is has two connected components $f^{-1}(V) = U_1 \cup U_2$. Pick three points $x, y, z$ such that $f(x), f(y), f(z) \in V$. Clearly, $x, y, z \in f^{-1}(V)$. Either two things can happen:

1. all z, y, and z are in the same component (they are all "close" to each other)
2. $x \in U_1$ but $y \in U_2$. Then z is going to be "close" to either of them but not both.

 

[kade]

This is how I visualize it. (For now I'm assuming everything are connected open sets.)

Consider metric spaces. Let $U = (0,1)$ and consider two points, say, $x_0 = \frac{1}{12}$ and $x_n = \frac{9}{10}$. Because U is open, I can find an open neighborhood of $x_0$, namely, $B(1/12, 1/12)$. Pick a point to the right inside of this ball: for example $x_1 = \frac{1}{7}$. Now find a new open neighborhood: $B(1/7, 1/7)$ and find a new point $x_2 = \frac{3}{14}$ and repeat.

What we end up with is a sequence $x_0, x_1, x_2, \ldots, x_n$ where we "walk" from $x_0$ to $x_n$ such that

1. every point is in U
2. neighborhoods are open subsets of U
3. successive neighborhoods overlap with each other.

Now replace neighborhood with "open subset" and you can frame that procedure in terms of topology.
But this doesn't work with closed sets. Try $U = [0,1]$ and pick $x_0 = 0$. There are no open neighborhoods of $x_0$ which are also subsets of U. So we can't do our walking. This is a difference between open and closed with regards to this (very non mathematical and very arbitrary) concept of "closeness".

But what happens when $f^{-1}(V)$ is not connected. Let's suppose that is has two connected components $f^{-1}(V) = U_1 \cup U_2$. Pick three points $x, y, z$ such that $f(x), f(y), f(z) \in V$. Clearly, $x, y, z \in f^{-1}(V)$. Either two things can happen:

1. all z, y, and z are in the same component (they are all "close" to each other)
2. $x \in U_1$ but $y \in U_2$. Then z is going to be "close" to either of them but not both.

Yea, the inverse image should be connected. But, according to the property of reserving closeness, I am thinking why not define the continuous function as this way:
It maps connected subsets in X onto connected subsets in Y; For any connected subset in Y, its inverse image in X is connected.

 

[pwsnafu]

For any connected subset in Y, its inverse image in X is connected.

Counter example: $f(x) = x^2$.

 

[pasmith]

Quoted from Wikipedia,

A function

between two topological spaces X and Y is continuous if for every open set V ⊆ Y, the inverse image

is an open subset of X.
How to comprehend this definition in a intuitive way?

Consider this definition of continuity at a point, which is the generalization to topological spaces of the definition of continuity at a point for metric spaces:

A function f: X \to Y is continuous at x \in X if and only if for every open neighbourhood V \subset Y of f(x) there exists an open neighbourhood U \subset X of x such that f(U) \subset V (equivalently, U \subset f^{-1}(V)).

This definition of local continuity is equivalent to the definition of global continuity in the sense that f is continuous at every x \in X if and only if f^{-1}(V) is open for every V \subset Y.

 

[mathwonk]

heres another nice equivalent version, if "cl" denotes closure, then for every subset W of the domain, f(cl(W)) is contained in cl(f(W)). Intuitively this means points infinitely close to W get mapped to points infinitely close to f(W).

 

[PeroK]

Quoted from Wikipedia,

A function

between two topological spaces X and Y is continuous if for every open set V ⊆ Y, the inverse image

is an open subset of X.
How to comprehend this definition in a intuitive way?

Note that for functions on metric spaces, this definition is equivalent to the normal $\epsilon - \delta$ definition. It's not hard to prove.

Intuitively, if a function has a discontinuity and its value "jumps" from $a$ to $b$ at the point x (let's say $f(x) = a < b$) then you can find an open set around $a$, whose preimage will be $(?, x]$. I.e. the preimage will include $x$, but no points immediately greater than $x$, hence not be open.

 

[WWGD]

Yea, the inverse image should be connected. But, according to the property of reserving closeness, I am thinking why not define the continuous function as this way:
It maps connected subsets in X onto connected subsets in Y; For any connected subset in Y, its inverse image in X is connected.

Like PwSnafu said, this is not necessarily true. For this you need the inverse function to be continuous (maybe there are weaker conditions than continuity, but not that I am aware of.) This gives you an interesting exercise you may want to try: show that a non-constant function that sends connected to connected is continuous.

 

[micromass]

This gives you an interesting exercise you may want to try: show that a non-constant function that sends connected to connected is continuous.

This is actually false. Did you want the OP to find a counterexample?

 

[kade]

Note that for functions on metric spaces, this definition is equivalent to the normal $\epsilon - \delta$ definition. It's not hard to prove.

Intuitively, if a function has a discontinuity and its value "jumps" from $a$ to $b$ at the point x (let's say $f(x) = a < b$) then you can find an open set around $a$, whose preimage will be $(?, x]$. I.e. the preimage will include $x$, but no points immediately greater than $x$, hence not be open.

This example makes sense. I got the picture now. Thank you.

 

[WWGD]

This is actually false. Did you want the OP to find a counterexample?

I wanted him/her to test it. I was not sure whether was true or not. Do you have a counter? Never mind, sorry for being lazy. EDIT: Still, the question I had in mind was a bit more difficult: not taking connected component to connected component, but assuming that the domain is itself connected and its image is connected.

 

[micromass]

I wanted him/her to test it. I was not sure whether was true or not. Do you have a counter? Never mind, sorry for being lazy. EDIT: Still, the question I had in mind was a bit more difficult: not taking connected component to connected component, but assuming that the domain is itself connected and its image is connected.

It seems like the topological sine function would do it, no?

 

[WWGD]

It seems like the topological sine function would do it, no?

Yes, I think so.

 

[micromass]

Some interesting results though:
1) A function $f:\mathbb{R}\rightarrow \mathbb{R}$ which preserves connected sets and compact sets is continuous. I don't know how much it generalizes to other metric spaces, but I know it is false in its most general form for topological spaces.

2) A function is continuous if and only if its graph is connected and locally connected.

 

[WWGD]

Yes, sorry, my idea was not as interesting/productive as I thought it would be.

 

[lavinia]

Yes, sorry, my idea was not as interesting/productive as I thought it would be.

Even though it is not true that a function is necessarily continuous if it preserves connectedness, I think your idea was interesting. It is true that the continuous image of a connected set is connected so this helps get some intuition for continuity. It is also true that the continuous image of a path connected set is path connected and this rules out the topologist's sine curve. Perhaps your intuition involved this idea.

The idea of continuity when applied to metric spaces boils down to Cauchy sequences. A function is continuous if whenever x_n converges to a point p then f(x_n) converges to f(p). One proves that the general definition, inverse images of open sets are open, is equivalent to this for metric spaces. It is then natural perhaps to ask such questions first about metric spaces.

Consider, for instance, the metric space that is itself a single Cauchy sequence e.g. X_n = 1/2^n and p = 0. The function 1/2^n -> n and 0 -> 0 preserves connectedness and path connectedness but is not continuous.

Still one might think that the image of any unbroken curve is always unbroken ought to suffice for some interesting spaces. What about path connected metric spaces? Can every Cauchy sequence be traversed by a continuous path?