Difference between revisions of "Continuity definitions are equivalent"

Revision as of 16:13, 13 February 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 16:17, 13 February 2015 (view source) Alec (Talk | contribs) m Newer edit →
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	<math>\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math>		<math>\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 f(a)\in 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.		Thus it is continuous at <math>x</math>, since <math>x</math> was arbitrary, it is continuous.
	{{Theorem|Topology}}		{{Theorem|Topology}}

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Revision as of 16:17, 13 February 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.