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Continuous function on metric spaces

Posted November 19th, 2007 by Structure

Let $ (X,d) $ be a metric space where X is a set and $ d:X\rightarrow [0,+\infty) $ is the distance, i.e.
d has properties
$ (i)\forall x,y\in X, d(x,y)\ge 0 $ and $ d(x,y)=0 $ if and only if $ x=y $.
$ (ii)\forall x,y\in X, d(x,y)=d(y,x) $
$ (iii)\forall x,y,z\in X, d(x,z)\le d(x,y)+d(y,z) $
There is a canonical topology on $ X $ namely $ \mathcal{T}_d $ such as $ (X,\mathcal{T}_d) $ a separated topological space. To describe the topology of $ (X,\mathcal{T}_d) $ it is useful to introduce a ball of center $ x\in X $ and radius r, i.e. the set

$$B(x,r)=\{y\in X|d(x,y)<r\}$$

Now a set $ G\subseteq X $ is open $ (G\in \mathcal{T}_d) $ if and only if for all $ x\in G $ there is a $ r>0 $ such as $ B(x,r)\subset G $.
Now we can add an equivalent condition to theorem "Continuous Function" about continuous function in this special case of topological metric space.
Let $ (X,d_X) $ and $ (Y,d_Y) $ metric spaces and $ f:\rightarrow Y $
Then f is continuous on X if and only if
$ (vi) \forall x\in X,\forall \epsilon>0, \exists \delta >0 $ such as $ \forall y\in X $ if $ d_X(x,y)<\delta, $ we have $ d_Y(f(x),f(y))<\epsilon $

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