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I'm a beginner in the course of Linear Algebra; please bear with me if the question seems too trivial.

The set of all continuous functions on interval $[0,1]$ is a vector space.

I have trouble in understanding this. What does a continuous function on $[0,1]$ mean? That the range lies within $[0,1]$?

For it to be a vector space, it needs to satisfy vector additivity.

Say, we take a vector with continuous function $f(t)=0.9$ which belongs to $V$. Another vector belonging to $V$ has continuous function $g(t)=0.8$.

For vector additivity, we add the elements of vector (in this case I'm considering only 1 element in the vector). Here the new vector would give us $0.8+0.9$ which us not in $[0,1]$. And yet this is a valid vector space.

I'm sure I'm missing something. I'm probably not able to understand what the question demands.


marked as duplicate by JMoravitz, Don Thousand, Xander Henderson, ancientmathematician, Sou Nov 1 '18 at 17:18

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  • $\begingroup$ Is it functions from $[0,1]$ or to $[0,1]$? If it is to $[0,1]$, this is not a vector space. $\endgroup$ – Don Thousand Nov 1 '18 at 16:56
  • $\begingroup$ Presumably we are talking about continuous real functions on $[0,1]$. A continuous real function on $[0,1]$ is a function $f$ with domain $[0,1]$ and codomain $\Bbb R$ satisfying the definition of what it means to be continuous. $\endgroup$ – JMoravitz Nov 1 '18 at 16:57
  • $\begingroup$ @RushabhMehta The question states "all continuous functions" $\endgroup$ – Shinjini Rana Nov 1 '18 at 16:58
  • $\begingroup$ @JMoravitz So is it the domain we are referring to when we say "on [0,1]"? $\endgroup$ – Shinjini Rana Nov 1 '18 at 16:59
  • $\begingroup$ It almost certainly is referring to the set of functions $\{ f: [0,1]\rightarrow \mathbb{R}: f \text{ is continuous on [0,1]} \}$ $\endgroup$ – Theo C. Nov 1 '18 at 16:59

A continuous function on $[0,1]$ is a function

$$f:[0,1] \to \mathbb{R}$$

which is continuous for every point in $[0,1]$. Since the sum and scalar multiples of continuous functions are also continuous (and addition is commutative) we have a vector space since there are additive inverses $(f-f=0)$ and a distinguished 0 element.

What might be throwing you is the fact that there is no finite basis for this vector space.


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