Let $a,b$ be two distinct points of a metric $(X,d)$ and $\alpha , \beta$ be any two given real number . Show that there exists a real valued continuous function $f$ on $X$ such that $f(a) = \alpha$ and $f(b) = \beta $.
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I suppose $(X,d)$ have to be complete. – Emilio Novati Feb 26 '15 at 14:52
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In such a case use $f(x)=d(a,x)$ – Emilio Novati Feb 26 '15 at 14:53
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Define $f : X \to \Bbb R$ by setting
$$f(x) = \alpha\frac{d(x,b)}{d(a,b)} + \beta \frac{d(a,x)}{d(a,b)}.$$
Since $x\mapsto d(x,b)$ and $x\mapsto d(a,x)$ are continuous and $d(a,b)\neq 0$ (as $a$ and $b$ are distinct), it follows that $f$ is continuous. Further, $f(a) = \alpha(1) + \beta(0) = \alpha$ and $f(b) = \alpha(0) + \beta(1) = \beta$.
kobe
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Hints: How about the function $$f(x) = \frac{\alpha \times d(x,b) + \beta \times d(x,a) }{d(x,a) +d(x,b)}.$$ for all $x \in X$.
User8976
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Notice that a metric space is normal and so the set $A=\{a,b\}$ is closed in $X$. Define $f:A\to\mathbb{R}$ by $f(a)=\alpha$ and $f(b)=\beta$. Then $f$ is clearly continuous. The Tietze Extension Theorem then guarantees that there is a continuous function $F:X\to\mathbb{R}$ such that $\left.F\right|_A=f$; that is to say, $F(a)=\alpha$ and $F(b)=\beta$.
rnrstopstraffic
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