Why is there no continuous bijection from (0,1] to R?

Why is there no continuous bijection from (0,1] to R?

I read somewhere that the reason is due to the fact that (0,1] is a connected set and the image of a connected set must be connected, so why can't we have a connected image?

The continuous image of $$(0,1)$$ must be connected, but by bijectivity this is $$\mathbb R$$ except the image of $$1$$. Removing any point from $$\mathbb R$$ makes it disconnected, a contradiction. Specifically, the two connected components of $$\mathbb R\backslash\{a\}$$ are $$(-\infty,a)$$ and $$(a,+\infty)$$.
Note that any injective continuous function from $$(0,1]$$ must be strict increase(or decrease). Hence $$f(1)$$ is either the maximal value or minimal value, but $$\mathbb{R}$$ don’t have maximal element or minimal element. This implies $$f$$ can’t be onto, if it is injective.
• @William: Yes. If you want to understand this properly you will need to look at IVT (intermediate value theorem), which guarantees the property Landau states. More specifically, given any continuous $f$ on any real interval $I$, there are no $a,b,c$ such that $a,b<c$ and $f(c)$ is strictly between $f(a)$ and $f(b)$. Once you prove this, you can apply it to $c=1$. Apr 11 '21 at 3:08