If $f : (0, \infty) \to \mathbb{R}^n$ is continuous on $[a, \infty) , \forall a>0$ then continuous on $(0, \infty)$

Question

How to prove/disprove

If $f : (0, \infty) \to \mathbb{R}^n$ is continuous on $[a, \infty) , \forall a>0$ then $f$ is continuous on $(0, \infty)$

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My try

Assume it is not continuous on $(0, \infty)$.

Let's denote a discontinuity point $x_0 \in (0,\infty)$. $\exists \epsilon >0$ s.t. $\forall \delta >0$, $|x - x_0| < \delta$ and $|f(x) - f(x_0)| \ge \epsilon $.

But since $f$ is continuous on $[x_0/2, \infty)$, it is contradiction.

This proof seems doubtful for me, but I cannot specify the doubts.

Is there anyone to clarify this?

Answer 1

It's true. You need to show continuity at any point $b$ say in $(0,\infty)$. Pick $a=b/2$ and from what you have assumed, you are there. Which perhaps justa neater way of writing what you have already done.

NB: if altered to 'then $f$ is continuous on $[0,\infty)$' it becomes false, which is perhaps what the question should have read in the first place. Similarly, if it had said 'then $f$ is uniformly continuous on $[0,\infty)$'.

Answer 2

Your proof is correct, but I think that it is simpler to say that $f$ cannot be discontinuous at a $x_0\in(0,\infty)$ since it is continuous at every point of $\left(\frac{x_0}2,\infty\right)$, to which $x_0$ belongs and which is an open subset of $(0,\infty)$.