Difference between revisions of "Notes:Continuous at a point"

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(Creating page with summary of problem)
 
(Problem: the pre-image need not be open)
 
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{{Todo|Attempt proof and explore}}
 
{{Todo|Attempt proof and explore}}
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:Indeed, the pre-image need not be open (otherwise it is about continuity everywhere, not just at a point). [[User:Boris|Boris]] ([[User talk:Boris|talk]]) 20:15, 23 March 2016 (UTC)

Latest revision as of 20:15, 23 March 2016

Problem

On the continuous map page there is a definition for continuity at a point.

Triggering definition

A map, [ilmath]f:X\rightarrow Y[/ilmath] between two topological spaces [ilmath](X,\mathcal{J})[/ilmath] and [ilmath](Y,\mathcal{K})[/ilmath] is continuous at [ilmath]x_0\in X[/ilmath] if:

  • [ilmath]\forall N\subseteq Y\text{ neighbourhood to }f(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0][/ilmath]

Question

I have noticed conflicting definitions of neighbourhood (see that page for details) and usually what is true for the "any set containing an open set containing the point" definition is true for the "any open set containing the point" definition. In that spirit I suspect the above definition if and only if:

  • [ilmath]\forall\mathcal{O}\in\mathcal{K}[f(x_0)\in\mathcal{O}\implies f^{-1}(\mathcal{O})\in\mathcal{J}\wedge x_0\in f^{-1}(\mathcal{O})][/ilmath]

HOWEVER

  1. This may be too strong, to say that for a function just continuous at a point EVERY open set containing the image of that point must have an open pre-image.

It is possible that it may be weakened to:

  • [ilmath]\forall\mathcal{O}\in\mathcal{K}[f(x_0)\in\mathcal{O}\implies f^{-1}(\mathcal{O})\text{ is a neighbourhood of }x_0][/ilmath]



TODO: Attempt proof and explore



Indeed, the pre-image need not be open (otherwise it is about continuity everywhere, not just at a point). Boris (talk) 20:15, 23 March 2016 (UTC)