10 - Limits and Continuity in Metric Spaces

Limit

Suppose that $(x_n{)}_{n=1}^{\infty }$ be a sequence of elements in metric space $(X,d)$

Let $x\in X$ then

$$x_n\to x\text{ or }\underset{n\to \infty }{\lim }{x}_n=x$$

If

$$\forall ϵ>0:\exists N\text{ such that }d(x_n,x)<ϵ\text{ for all }n\ge N$$

Continuity

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces

Function $f:X\to Y$ is continuous at $a\in X$ if

$$\forall ϵ>0:\exists \delta >0\text{ such that }\forall x\in X:d_X(a,x)<\delta ⟹d_Y(f(x),f(a))<ϵ$$

So $f$ is continuous if $f$ is continuous at every $a\in X$

Uniform Continuity

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces

Function $f:X\to Y$ is uniformly continuous at $a\in X$ if

$$\forall ϵ>0:\exists \delta >0\text{ such that }\forall x,y\in X:d_X(x,y)<\delta ⟹d_Y(f(x),f(y))<ϵ$$

So $f$ is continuous if $f$ is continuous at every $a\in X$

Applying Limits through functions lemma

Let $f:X\to Y$ be a function between metric spaces then

$$f\text{ is continuous}⟺\text{ for any }(x_n{)}_{n=1}^{\infty }\text{ with }\underset{n\to \infty }{\lim }=a\text{ means }\underset{n\to \infty }{\lim }f(x_n)=f(a)$$

Proof

Suppose that $f$ is continuous at $a$
Let $ϵ>0$ then exists $\delta >0$ such that

$$\text{ for all }x\in X\text{ with }d(x,a)<\delta \text{ then }d(f(x)),f(a))<ϵ$$

Let $(x_n{)}_{n=1}^{\infty }$ be a sequence with limit $\alpha$

By definition of limit then exists $N>0$ such that

$$d(a,x_n)<\delta \text{ for all }n\ge N$$

However for all $n\ge N$ then

$$d(f(a),f(x_n))<ϵ\text{ so that }\underset{n\to \infty }{\lim }f(x_n)=f(a)$$

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Other direction just show contrapositive
Suppose $f$ is not continuous at $a$ hence

$$\exists ϵ>0:\forall \delta >0:\exists x\in X:d(x,a)<\delta ⟹d(f(x),d(a))\ge ϵ$$

Taking $\delta =\frac{1}{n}$ then there is some $x_n\in X$ with

$$d(x_n,a)<\frac{1}{n}⟹d(f(x_n),f(a))\ge ϵ$$

Hence

$$\underset{n\to \infty }{\lim }{x}_n=a\text{ but }\underset{n\to \infty }{\lim }f(x_n)\ne f(a)$$