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

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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, SouNov 1 '18 at 17:18

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

## 1 Answer

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.