calculus - Spivak Continuity problem

Suppose that $f$ is continuous from the right at $a$ and there exists $\delta>0$ s.t if $a0$. Prove that $f(a)\geqslant0$.

In Spivak he proves this theorem in ch.7 and so I used it in this proof:

If $f$ is continuous on $[a,b]$ and $f(a)<0

My attempt at a proof using this theorem and contradiction.

Suppose $f(a)<0$. If f is continuous from the right, it is continuous on the interval $a0. Thus $f(a)\geqslant0$

My questions are can I extend the theorem spivak gives to open intervals? And since I'm only given that the function $f$ is continuous from the right does my contradiction using $f(a)<0$ work since I don't necessarily know what happens to the function past $a$? Thanks.

Answer

Your proof uses the fact that $f$ is continuous in $a

Do not use non-trivial theorems to prove trivial results. Here is how one can prove the result in question. Let $f(a) <0$. Since $f$ is continuous at at $a$ from right there is a $\delta_{1}>0$ such that $$|f(x) - f(a)|<-\frac{f(a)} {2}$$ whenever $a\leq x

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I add an easier, informal but rigorous version. Since $f$ is continuous from right at $a$, values of $f$ to the right of and near $a$ are near to $f(a) $ and if $f(a) $ is negative then these values of $f$ are also negative (numbers close to a negative number are negative). This contradicts that values of $f$ to the right of and near $a$ are positive.

Ideally one should be able to think informally as above and if needed produce the formal version instantly. Most of the proofs in analysis deal with such informal thinking and their translation into their boring formal counterparts.