Difference between revisions of "Continuous map"
m |
m |
||
| Line 1: | Line 1: | ||
| + | {{Refactor notice}} | ||
| + | :: '''Note: ''' there are a few different conditions for continuity, there's also continuity at a point. This diagram is supposed to show how they relate to each other. | ||
| + | |||
| + | {| class="wikitable" border="1" | ||
| + | |- | ||
| + | | style="font-size:1.2em;" | | ||
| + | {{MM|1=\begin{xy}\xymatrix{ | ||
| + | \text{Continuous} \ar@2{<->}[d]_-{\text{claim }1} \ar@2{<.>}[drr] & & \\ | ||
| + | {Continuous at x0(neighbourhood) } \ar@2{<->}[rr]_{\text{claim }2} & & {Continuous at x0(sequential) } | ||
| + | }\end{xy} }} | ||
| + | | '''Note that: ''' | ||
| + | * All arrow denote logical [[implies]], or "if and only if" | ||
| + | * Dotted arrows show immediate results of the claims on this page | ||
| + | |- | ||
| + | ! Overview | ||
| + | ! Key | ||
| + | |} | ||
| + | |||
| + | __TOC__ | ||
| + | ==Definition== | ||
| + | Given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}} we say that a [[map]], {{M|f:X\rightarrow Y}} is continuous if<ref name="KMAPI">Krzysztof Maurin - Analysis - Part 1: Elements</ref>: | ||
| + | * {{M|\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}]}} | ||
| + | That is to say: | ||
| + | * The [[pre-image]] of every set open in {{M|Y}} under {{M|f}} is open in {{M|X}} | ||
| + | ==Continuous at a point== | ||
| + | Again, given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}}, and a point {{M|x_0\in X}}, we say the [[map]] {{M|f:X\rightarrow Y}} is ''continuous at {{M|x_0}}'' if<ref name="KMAPI"/>: | ||
| + | * {{M|\forall N\subseteq Y}}[[neighbourhood|{{M|\text{ neighbourhood to } }}]]{{M|T(x_0)[f^{-1}(N)\text{ is a neighbourhood of }x_0]}} | ||
| + | {{Begin Inline Theorem}} | ||
| + | '''UNPROVED: '''{{Note|I suspect that this is the same as {{M|\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})]}} - this is basically the same just on open sets instead}} | ||
| + | {{Begin Inline Proof}} | ||
| + | {{Todo|{{Note|Investigate and prove the highlighted claim}}}} | ||
| + | (Leave any notes to self here) | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | ===Claim 1=== | ||
| + | {{Begin Theorem}} | ||
| + | Claim: The [[mapping]] {{M|f}} is continuous {{M|\iff}} it is continuous at every point | ||
| + | {{Begin Proof}} | ||
| + | {{Todo|Do this}} | ||
| + | {{End Proof}}{{End Theorem}} | ||
| + | ==Sequentially continuous at a point== | ||
| + | Given two [[topological space|topological spaces]] {{M|(X,\mathcal{J})}} and {{M|(Y,\mathcal{K})}}, and a point {{M|x_0\in X}}, a [[function]] {{M|f:X\rightarrow Y}} is said to be ''continuous at {{M|x_0}}'' if<ref name="KMAPI"/>: | ||
| + | * {{MM|1=\forall (x_n)_{n=1}^\infty\left[\lim_{n\rightarrow\infty}(x_n)=x\implies\lim_{n\rightarrow\infty}(f(x_n))=f(x)\right]}} (Recall that {{M|1=(x_n)_{n=1}^\infty}} denotes a [[sequence]], see [[Limit (sequence)]] for information on limits) | ||
| + | ===Claim 2=== | ||
| + | {{Begin Theorem}} | ||
| + | Claim: {{M|f}} is continuous at {{M|x_0}} using the neighbourhood definition {{M|\iff}} it is continuous at {{M|x_0}} using the sequential definition | ||
| + | {{Begin Proof}} | ||
| + | {{Todo|Fill this out}} | ||
| + | {{End Proof}} | ||
| + | {{End Theorem}} | ||
| + | ==References== | ||
| + | <references/> | ||
| + | {{Definition|Topology|Metric Space}} | ||
| + | |||
| + | =Old page= | ||
==First form== | ==First form== | ||
The first form: | The first form: | ||
Revision as of 04:22, 22 November 2015
- Note: there are a few different conditions for continuity, there's also continuity at a point. This diagram is supposed to show how they relate to each other.
|
|
Note that:
|
| Overview | Key |
|---|
Contents
[hide]Definition
Given two topological spaces (X,J) and (Y,K) we say that a map, f:X→Y is continuous if[1]:
- ∀O∈K[f−1(O)∈J]
That is to say:
- The pre-image of every set open in Y under f is open in X
Continuous at a point
Again, given two topological spaces (X,J) and (Y,K), and a point x0∈X, we say the map f:X→Y is continuous at x0 if[1]:
- ∀N⊆Y neighbourhood to T(x0)[f−1(N) is a neighbourhood of x0]
UNPROVED: I suspect that this is the same as ∀O∈K[f(x0)∈O⟹f−1(O)∈J∧x0∈f−1(O)] - this is basically the same just on open sets instead
Claim 1
Sequentially continuous at a point
Given two topological spaces (X,J) and (Y,K), and a point x0∈X, a function f:X→Y is said to be continuous at x0 if[1]:
- ∀(xn)∞n=1[lim (Recall that (x_n)_{n=1}^\infty denotes a sequence, see Limit (sequence) for information on limits)
Claim 2
Claim: f is continuous at x_0 using the neighbourhood definition \iff it is continuous at x_0 using the sequential definition
References
- ↑ Jump up to: 1.0 1.1 1.2 Krzysztof Maurin - Analysis - Part 1: Elements
Old page
First form
The first form:
f:A\rightarrow B is continuous at a if:
\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon (note the implicit \forall x\in A)
Second form
Armed with the knowledge of what a metric space is (the notion of distance), you can extend this to the more general:
f:(A,d)\rightarrow(B,d') is continuous at a if:
\forall\epsilon>0\exists\delta>0:d(x,a)<\delta\implies d'(f(x),f(a))<\epsilon
\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))
In both cases the implicit \forall x is present. Basic type inference (the B_\epsilon(f(a)) is a ball about f(a)\in B thus it is a ball in B using the metric d')
Third form
The most general form, continuity between topologies
f:(A,\mathcal{J})\rightarrow(B,\mathcal{K}) is continuous if
\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J} - that is the pre-image of all open sets in (A,\mathcal{J}) is open.
