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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.

Overview	Key
$\begin{xy}\xymatrix{ \text{Continuous} \ar@2{<->}[d]_-{\text{claim }1} \ar@2{<.>}[drr] & & \\ {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(neighbourhood)}\end{array} } \ar@2{<->}[rr]_{\text{claim }2} & & {\begin{array}{lr}\text{Continuous at }x_0\\ \text{(sequential)}\end{array} } }\end{xy}$	Note that: All arrow denote logical implies, or "if and only if" Dotted arrows show immediate results of the claims on this page

Definition

Given two topological spaces $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ we say that a map, $f:X\rightarrow Y$ is continuous if^[1]:

$\forall\mathcal{O}\in\mathcal{K}[f^{-1}(\mathcal{O})\in\mathcal{J}]$

That is to say:

The pre-image of every set open in $Y$ under $f$ is open in $X$

Continuous at a point

Again, given two topological spaces $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ , and a point $x_0\in X$ , we say the map $f:X\rightarrow Y$ is continuous at $x_0$ if^[1]:

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

Claim 1

[Expand]
Claim: The mapping $f$ is continuous $\iff$ it is continuous at every point

Sequentially continuous at a point

Given two topological spaces $(X,\mathcal{J})$ and $(Y,\mathcal{K})$ , and a point $x_0\in X$ , a function $f:X\rightarrow Y$ is said to be continuous at $x_0$ if^[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 $(x_n)_{n=1}^\infty$ denotes a sequence, see Limit (sequence) for information on limits)

Claim 2

[Expand]
Claim: $f$ is continuous at $x_0$ using the neighbourhood definition $\iff$ it is continuous at $x_0$ using the sequential definition

Proof: neighbourhood $\implies$ sequential:

Let

$(x_n)_{n=1}^\infty$ be given, and that

$(x_n)\rightarrow x$ - we wish to show that

$\lim_{n\rightarrow\infty}(f(x_n))=f(x)$

Let

$\epsilon > 0$ be given.

By hypothesis, as

$B_\epsilon(f(x))$ is a neighbourhood of

$f(x)$ then

$f^{-1}(B_\epsilon(f(x)))$ is a neighbourhood to

$x$

So

$\exists\delta>0[B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))$

By the implies-subset relation if $B_\delta(x)\subseteq f^{-1}(B_\epsilon(f(x)))$ then $a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))$

Choose

$N\in\mathbb{N}$ such that

$n>N\implies d_1(x_n,x)<\delta$ (which we can do because

$(x_n)\rightarrow x$ )

Note that $d_1(x_n,x)<\delta\implies x_n\in B_\delta(x)$

So

$x_n\in B_\delta(x)\implies x_n\in f^{-1}(B_\epsilon(f(x)))$

But if $a\in f^{-1}(B)$ then $f(a)\in B$ as $f^{-1}(B)$ contains exactly the things which map to an element of $B$ under $f$

So

$f(x_n)\in B_\epsilon(f(x))$

But

$f(x_n)\in B_\epsilon(f(x))\iff d_2(f(x_n),f(x))<\epsilon$

This completes the proof

we have shown that given an $\epsilon > 0$ that there exists an $N$ such that for $n>N$ we have $d_2(f(x_n),f(x))<\epsilon$ which is exactly the definition of $(f(x_n))\rightarrow f(x)$

Proof: sequential $\implies$ neighbourhood:

TODO: This - See Analysis I - Maurin page 44 if stuck

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.

Equivalence of definitions

Continuity definitions are equivalent

[KMAPI-1] {Jump up to: 1.0} ^1.1 ^1.2 Krzysztof Maurin - Analysis - Part 1: Elements

[1]

@@ Line 26: / Line 26: @@
 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|f(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}}

Difference between revisions of "Continuous map"

Revision as of 12:19, 11 May 2016

Contents

Definition

Continuous at a point

Claim 1

Sequentially continuous at a point

Claim 2

References

Old page

First form

Second form

Third form

Equivalence of definitions

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