Continuous map

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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.

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

Definition

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

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

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:

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

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UNPROVED: I suspect that this is the same as $\forall\mathcal{O}\in\mathcal{K}[f(x_0)\in\mathcal{O}\implies f^{-1}(\mathcal{O})\in\mathcal{J}\wedge x_0\in f^{-1}(\mathcal{O})]$ - this is basically the same just on open sets instead

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TODO: Investigate and prove the highlighted claim

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(Leave any notes to self here)

Claim 1

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

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• A quick proof I did on some scrap - click for full version

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TODO: Write up proof

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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:

• $$\forall (x_n)_{n=1}^\infty\left[\lim_{n\rightarrow\infty}(x_n)=x\implies\lim_{n\rightarrow\infty}(f(x_n))=f(x)\right]$$ (Recall that $(x_n)_{n=1}^\infty$ denotes a sequence, see Limit (sequence) for information on limits)

Claim 2

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

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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:

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

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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