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If a function $f$ is continuous at every $c$ point in $[a, b]$ then we have for every $\epsilon$ we can find a $\delta$ such that whenever $|x-c|<\delta$, we have $|f(x)-f(c)|<\epsilon$.

Now, in this definition, $c$ varies in the interval $[a, b]$, and also $x$ varies in $[a, b]$, then what is the difference between $x$ and $c$?

Blue
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user679770
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    $c$ is a fixed point and $x$ is a variable. – Kavi Rama Murthy Jul 31 '19 at 08:52
  • We have the def of continuity of $f$ at point $c$, where $c$ is a specific point, and then we have a new def of continuity in an interval $[a,b]$ where the previous def of continuity at a single point is "generalized" to every point in the interval. That's all... – Mauro ALLEGRANZA Jul 31 '19 at 08:54
  • As said in the previous comments: $\bullet$ $f$ is continuous at $c$ if for every $\epsilon>0$ there exists $\delta>0$ such that for every $x$ satisfying $|x-c|<\delta$ ($x$ varies, $c$ is fixed) it holds $|f(x)-f(c)|<\epsilon$. $\bullet$ $f$ is continuous on $(a.b)$ if for every $c\in (a,b)$, $f$ is continuous at $c$. – Surb Jul 31 '19 at 08:58
  • You find $δ$ for given $c$ and $ϵ$. If $δ$ only depends on $ϵ$, so that $c$ and $x$ have equal rank (in the sense of placement), then you are dealing with uniform continuity. – Lutz Lehmann Jul 31 '19 at 09:22

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A function $f$ is continuous on $[a,b]$ if for all $c\in[a,b]$ and all $\varepsilon$ there exists a $\delta$ (dependent of both $c\in[a,b]$ and $\varepsilon$) such that for all $x$ with $|x-c|<\delta$ we have $|f(x)-f(c)|<\varepsilon$.

The order of when you pick $c$, $\varepsilon$, $\delta$ and $x$ matters. What you are thinking about is uniform continuity: A function $f$ is continuous on $[a,b]$ if for all $\varepsilon$ there exists a $\delta$ (dependent only on $\varepsilon$ such that for all $c,x\in[a,b]$ with $|x-c|<\delta$ we have $|f(x)-f(c)|<\varepsilon$. In this case $\delta$ is independent of $c$.

Floris Claassens
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Working with some function $f:[a,b]\to\mathbb R$ and using quantors we have:$$f\text{ is continuous at }c\iff\forall \epsilon>0\;\exists \delta>0\;\forall x\in[a,b]\;\left[|x-c|\leq\delta\implies |f(x)-f(c)|<\epsilon\right]$$Note that here $c$ is a free variable and $x$ is a bound variable (i.e. connected with a quantor).

We are dealing here with a statement about $f$ and $c$.

Further:$$f\text{ is continuous }\iff\forall c\in[a,b]\;\left[f\text{ is continuous at }c\right]\iff$$$$\forall c\in[a,b]\;\forall \epsilon>0\;\exists \delta>0\;\forall x\in[a,b]\;\left[|x-c|\leq\delta\implies |f(x)-f(c)|<\epsilon\right]$$ Note that here $c$ and $x$ are both bound variables and that the meaning of the statement on RHS does not change if the variables $c$ and $x$ are switched.

We are dealing here with a statement about $f$.

drhab
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