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A path in a topological space X is a continuous function from the closed unit interval [0, 1] into X.

What happens when the topological space is something more simple, for example given $X = \{ 1, 2, 3, 4\},$ consider the topology $\tau = \{ \varnothing, \{ 2 \}, \{1, 2\}, \{2, 3\}, \{1, 2, 3\}, X \}$ of six subsets of $X$.

If I now have a path in the example topological space $(X,\tau)$, what is the respective continuous function? Does it map from time to subsets? How can it be continuous if it jumps discretely from subset to subset?

The concept is very intuitive in e.g. $\mathbb{R}^2$, but not so clear what is meant in this elementary case.

apg
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  • A path in your example $X$ would be a function taking values in $X$ not subsets of $X$. I suggest you try some examples to see whether you can find a continuous function that is not constant. It may be of interest that if you give any set $A$ the indiscrete topology (just two open sets), then any function to $A$ is continuous, but your example is more complicated. – Rob Arthan Apr 27 '22 at 12:58
  • Is the usual definition of continuous used? Small movements in time mean small movements in the space? What is a small movement in a set with two elements? – apg Apr 27 '22 at 13:28
  • Who defines continuity in terms of “small movements”? – José Carlos Santos Apr 27 '22 at 13:31
  • The $\epsilon-\delta$ definition? – apg Apr 27 '22 at 13:33
  • In the abstract setting "small movements cause small changes" requires some pretty vigorous handwaving to make sense of. – Ian Apr 27 '22 at 14:29
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    @apkg: The definition to use is that $f : X \to Y$ is continuous if $f^{-1}[A]$ (the inverse image of $A$) is open in $X$ whenever $A$ is open in $Y$. This generalises the $\epsilon$-$\delta$ definition to arbitrary topological spaces. – Rob Arthan Apr 27 '22 at 15:51

2 Answers2

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$f:[0, 1]\to X$ defined by $$f(x)=\begin{cases} 1\quad \quad \text{$x\in (0,\frac{1}{6})$} &\\ 2\quad \quad \text{$x\in (\frac{1}{5},\frac{1}{4})$}&\\ 3\quad \quad \text{$x\in (\frac{1}{3},\frac{1}{2})$}&\\ 4\quad \quad \text{otherwise} \end{cases}$$

Then $f$ is continuous map as pre-image of every open set is open.

$f^{-1}(\{2\}) =(\frac{1}{5},\frac{1}{4})$

$f^{-1}(\{1,2\}) =(0,\frac{1}{6})\cup (\frac{1}{5},\frac{1}{4}) $

$f^{-1}(\{2,3\}) = (\frac{1}{5},\frac{1}{4})\cup (\frac{1}{3},\frac{1}{2}) $

$\begin{align} f^{-1}(\{1,2,3\}) &= (0,\frac{1}{6})\cup(\frac{1}{5},\frac{1}{4})\cup (\frac{1}{3},\frac{1}{2}) \end{align}$

$f^{-1}(X) =[0, 1]$

$f^{-1}(\emptyset ) =\emptyset$

Sourav Ghosh
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  • Thank you. So the idea of continuity is that the pre-image of every open set is open, and then you just use a function set up in pieces, mapping blocks of time to elements of the set $X$. – apg Apr 27 '22 at 13:37
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    The key thing here is that $4$ is "stuck to everything", since there is no open set other than $X$ itself that contains it. So you can pass through it on a continuous path to anywhere else. – Ian Apr 27 '22 at 13:39
  • So is this then a path from 0 to 4? – apg Apr 27 '22 at 13:47
  • In the end this is a path from 1 to 4, though it goes through 4 several times along the way. – Ian Apr 27 '22 at 14:19
  • Sorry, literally as written this is a path from $4$ back to $4$, but you could change $(0,1/6)$ to $[0,1/6)$ and it would be a continuous path from $1$ to $4$. But explaining why that makes sense requires talking about the subspace topology. – Ian Apr 27 '22 at 14:41
  • What about a path from 1 to 2? Does every element of the topology have to have an open pre-image in [0,1]? Or can the path just traverse a subspace? – apg Apr 27 '22 at 15:31
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    @apkg You could have an open set that's just never hit, and that would be fine because $\emptyset$ is open. In particular a constant function is always continuous regardless of the topologies on the domain and codomain. – Ian Apr 27 '22 at 15:47
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Topology involves the abstraction of intuitively clear notions, which is what you're butting up against. The "small changes in input lead to small changes in output" intuition regarding continuity is often useful, but it's much cleaner to think strictly in topological terms here. The $\epsilon$-$\delta$ definition of continuity requires a notion of distance, which means you have to work with metric spaces. You haven't defined either of your spaces as metric spaces.

Try using the abstract, topological definition of continuity: a function $f$ is continuous if the preimage of an open set (in the range topology) is open (in the domain topology).

Note that "open" is a relative concept, i.e. it depends on the specific topology being considered. To drive this point home, consider the function $g:[0, 1]\to X$ defined by $$g(x)=\begin{cases} 1\quad \quad \text{$x\in [0,\frac{1}{4})$} &\\ 2\quad \quad \text{$x\in [\frac{1}{4},\frac{1}{2})$}&\\ 3\quad \quad \text{$x\in [\frac{1}{2},\frac{3}{4})$}&\\ 4\quad \quad \text{$x\in [\frac{3}{4},1]$} \end{cases}$$

where we think of $[0, 1]$ as having the subspace topology when $\mathbb R$ has the lower limit topology. In this case $g$ is a path, i.e. a continuous function, although it would not be continuous if we had initially thought of $\mathbb R$ as having its regular topology, as was done in the accepted answer.

apg
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Novice
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    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – KBS Apr 27 '22 at 14:18
  • @KBS I count four or five questions. The first one is answered within the post, while the rest are ill-formed enough that I see no purpose in attempting to answer them directly. The accepted answer does not actually appear to answer any of the questions asked. In sum, I don't find this criticism persuasive. – Novice Apr 27 '22 at 16:31
  • This is a template message from the review process as your "answer" has been flagged because it was considered to be not an answer. If some questions are ill-formulated ask for a correction in the comments of the question. If you are only partially answering the question(s), mention it clearly in the post. – KBS Apr 27 '22 at 16:40
  • Thank you, that is clear and directly answers my question. What I needed was the notion of topological continuity. Thanks again. – apg Apr 27 '22 at 19:06