2.3 Limit and Continuity of a Function in \((X,d)\)
Definition 2.17. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and \(x_0\in X\). The function \(f:X\rightarrow Y\) is said to approach a limit \(l\in Y\) as \(x\) approaches \(x_0\) if given \(\varepsilon >0, \exists \delta >0\ni d_Y(f(x),l)<\varepsilon \) whenever \(0<d_X(x,x_0)<\delta \).
Definition 2.18. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. Let \(f:X\rightarrow Y\) be a function. The function \(f\) is said to be continuous at \(x_0\in X\) if given \(\varepsilon >0,\exists \delta >0\ni d_Y(f(x),f(x_0))<\varepsilon \) whenever \(d_X(x,x_0)<\delta \).
Definition 2.19. Let \((X,d)\) be a metric space, if \(x_0\in X\) and \(r>0\), then the open ball or (open sphere) is the set \(B_r(x_0)=\{x\in X:d(x,x_0)<r\}\) or \(S_r(x_0)=\{x\in X:d(x,x_0)<r\}.\)
Example 2.20. Let \(X=\mathbb {R}\) with the usual metric, then \(B_r(x_0)=\{x\in \mathbb {R}:|x-x_0|<r\}\) or as an interval \((x_0-r,x_0+r)\).
Theorem 2.21. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and let \(f:X\rightarrow Y\) be a function, then \(f\) is continuous if and only if \(f^{-1}(G)\) is an open set in \(X\) whenever \(G\) is an open set in \(Y\).
Proof. Let \(f\) be continuous and \(G\) be an open set in \(Y\). Then to prove that \(f^{-1}(G)\) is open in \(X\). Choose an
arbitrary point \(x\in f^{-1}(G)\), then \(f(x)\in G\). Now, since \(G\) is open in \(Y\), it follows that \(B_r(f(x))\subseteq G\qquad (1)\). Since \(f\) is continuous , it follows
that
\(f(B_{\delta }(x))\subseteq B_r(f(x))\qquad (2).\hspace {0.2cm}\) Using \((1)\) and \((2)\) to get
\[f(B_{\delta }(x))\subseteq G\]
\[B_{\delta }(x)\subseteq f^{-1}(G)\]
Thus we’ve found an open ball for any \(x\in f^{-1}(G)\). Thus \(f^{-1}(G)\) is open in \(X\).
Conversely, Let \(G\) be an open set in \(Y\) and \(f^{-1}(G)\) be also open. Then to prove that \(f\) is continuous.
Consider the open set \(G=B_r(f(x))\). Then \(\exists \) some \(\delta >0\ni B_{\delta }(x)\subseteq f^{-1}(G)\). Therefore \(\hspace {0.2cm} f(B_{\delta }(x))\subseteq G=B_r(f(x)).\hspace {0.2cm}\) This shows continuity.
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Definition 2.22. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces and let \(f:X\rightarrow Y\). We say that \(f\) is uniformly continuous if for every \(\varepsilon >0, \exists \delta >0\ni d_Y(f(x_1),f(x_2))<\varepsilon \) whenever \(d_X(x_1,x_2)<\delta \) where \(x_1,x_2\in X\).
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