Difference between revisions of "Continuity definitions are equivalent"

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−	{{Todo|Theorem B16 in Introduction to topological manifolds is a good guide to structure}}	+
		+	==Statement==
		+	The definitions of continuity for a function <math>f:(X,d)\rightarrow(Y,d')</math> from one [[Metric space|metric space]] to another is the same as <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> being continuous (where the topologies are those [[Topology induced by a metric|induced by the metric]] are the same, that is
		+	# <math>\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))</math>
		+	# <math>\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}</math>
		+
		+	==Proof==
		+	===<math>\implies</math>===
		+	Suppose <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> is continuous.
		+
		+	Let <math>V\in\mathcal{K}</math> - that is <math>V</math> is [[Open set|open]] within <math>Y</math>
		+
		+	Let <math>x\in f^{-1}(V)</math> be given.
		+
		+	Then because <math>V</math> is open, <math>\exists\epsilon>0</math> such that <math>B_\epsilon(f(x))\subset V</math> (note that <math>f(x)\in V</math> by definition of where we choose x from).
		+
		+	But by continuity of <math>f</math> we know that <math>\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V</math>
		+
		+	Thus <math>B_\delta(x)\subset f^{-1}(V)</math> (as for all <math>a</math> in the ball, the thing <math>f</math> maps it to is in the ball of radius <math>\epsilon</math> about <math>f(x)</math>).
		+
		+	Since <math>x</math> was arbitrary we have <math>\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)</math>, thus <math>f^{-1}(V)</math> is open.
		+
		+	===<math>\impliedby</math>===
		+	Choose any <math>x\in X</math>
		+
		+	Let <math>\epsilon>0</math> be given.
		+
		+	As <math>B_\epsilon(f(x))</math> is an open set, the hypothesis implies that <math>f^{-1}(B_\epsilon(f(x)))</math> is open in <math>X</math>
		+
		+	Since <math>x\in f^{-1}(B_\epsilon(f(x)))</math> and <math>f^{-1}(B_\epsilon(f(x)))</math> is open, it is a neighborhood to all of its points, that means
		+
		+	<math>\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math>
		+
		+	''Note:'' we have now shown that <math>\forall\epsilon>0\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math>
		+
		+	Using the [[Implies and subset relation|implies and subset relation]] we see <math>a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))</math>
		+
		+	Or just <math>a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))</math>
		+
		+	Thus it is continuous at <math>x</math>, since <math>x</math> was arbitrary, it is continuous.
		+
		+	{{Theorem Of|Topology}}

---

Latest revision as of 07:20, 27 April 2015

Statement

The definitions of continuity for a function

$$f:(X,d)\rightarrow(Y,d')$$ from one metric space to another is the same as $$f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$$ being continuous (where the topologies are those induced by the metric are the same, that is

1. $$\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))$$
2. $$\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}$$

Proof

$$\implies$$

Suppose

$$f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})$$ is continuous.

Let

$$V\in\mathcal{K}$$ - that is $$V$$ is open within $$Y$$

Let

$$x\in f^{-1}(V)$$ be given.

Then because

$$V$$ is open, $$\exists\epsilon>0$$ such that $$B_\epsilon(f(x))\subset V$$ (note that $$f(x)\in V$$ by definition of where we choose x from).

But by continuity of

$$f$$ we know that $$\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V$$

Thus

$$B_\delta(x)\subset f^{-1}(V)$$ (as for all $$a$$ in the ball, the thing $$f$$ maps it to is in the ball of radius $$\epsilon$$ about $$f(x)$$ ).

Since

$$x$$ was arbitrary we have $$\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)$$ , thus $$f^{-1}(V)$$ is open.

$$\impliedby$$

Choose any

$$x\in X$$

Let

$$\epsilon>0$$ be given.

As

$$B_\epsilon(f(x))$$ is an open set, the hypothesis implies that $$f^{-1}(B_\epsilon(f(x)))$$ is open in $$X$$

Since

$$x\in f^{-1}(B_\epsilon(f(x)))$$ and $$f^{-1}(B_\epsilon(f(x)))$$ is open, it is a neighborhood to all of its points, that means

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

Note: we have now shown that

$$\forall\epsilon>0\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))$$

Using the implies and subset relation we see

$$a\in B_\delta(x)\implies a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))$$

Or just

$$a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))$$

Thus it is continuous at

$$x$$ , since $$x$$ was arbitrary, it is continuous.