Talk:Continuous Image of Separable Space is Separable

Being unable to come up with a proof, I started to suspect some premises were missing. Some web search revealed that the condition missing is likely that $f$ be surjective. Most sources seem to state that a 'continuous image of a separable space is separable'; this is wholly different from the statement currently up. I suggest looking at 'Counterexamples' to determine how to reformulate the result (I can't imagine that stuff as basic as this went wrong in that classic). --Lord_Farin 15:56, 22 July 2012 (UTC)

You'd be surprised. Check out this page: Category:Mistakes/Counterexamples in Topology ... Seebach and Steen are not the messiahs of Topology, they're very naughty boys.

I will recheck this in due course - but in the meantime let this page stand with the "questionable" template. --prime mover 16:32, 22 July 2012 (UTC)

... I've checked it, and it seems you're right. To quote:

...a property is called a continuous, open, or closed invariant if any continuous (respectively open, closed) image of a space possessing the property also possesses the property. Both separability and compactness are continuous invariants.

I'm going to have to think about the difference between what I put and "a continuous image of a separable space is separable." The same applies to the "compactness" which I already put an "error" page in place.

Feel free to put all this right, if you're on top of this - I'm missing a subtlety and my head's somewhere else right now. --prime mover 16:44, 22 July 2012 (UTC)

Thing is, a function is by definition surjective onto its range. If a function from $(W, \mathscr S)$ to $(X, \mathscr T)$ is not surjective onto $X$ itself then it is surjective onto its range $\subset X $ and there is subspace topology applicable to the range, being {range $ \cap T: T \in \mathscr T$}. So I finished the proof. Tom Collinge 17 feb 2016.

Yep, good stuff. You know you can add your name and timestamp to a chat post by pressing the button at the top of the wiki editor? It's the one second from the right with a squiggle on it (meant to be a signature). It adds two hyphens and four tildes to the edit at the point you press it. The four tildes get converted into your user name (with a link to your user page) and the time you committed the page. --prime mover (talk) 14:39, 17 February 2016 (UTC)

I have the impression it's most easily done by defining 'continuous invariant' and its cousins. The problem in general is that we can take an inclusion mapping to some arbitrary space (eg. a disjoint union) which behaves as bad as we can imagine; restricting to images is effectively imposing surjectivity; it is clear that this yields a more relaxed condition, as (most) inclusions are not in the picture. I have no source works on topology, so I'll leave it for now; just cruising through the stubs, seeing what I may prove. --Lord_Farin 16:51, 22 July 2012 (UTC)

Cruising through stubs is something I ought to get back to doing some time - I may find I have the experience to complete some of it which I might not have had before. But I'm still working at sorting out my source works. --prime mover 17:13, 22 July 2012 (UTC)

Rewritten proof

I have rewritten the proof so that it does not refer to $f$ being subjective, but focuses instead on the property holding in the image set of $f$. This now fits the form of the definition of the concept, and does not restrict us to surjections. --prime mover (talk) 08:44, 6 January 2018 (EST)