# Topology -- Continuity and the induced topology.

Here is my question.

Let $X=Y=\Bbb R$, with the usual topology. Let $A=[0,1]$ and topologize $A$ with the induced topology from $X$.

Does there exist a continuous function from the topological space $A$ onto $Y$?

Why or why not?

I'd argue No, since we'd be going from a closed to open set. But I also thought that any subset of a topology is an open subset.

Any help would be appreciated.

• Hint: All of $\mathbb{R}$ is a closed set (also an open set; it's clopen). The tan and arctangent functions have the required type of behavior, just rescale them as needed (domain restrict the tan function to make it invertible). Commented May 9, 2016 at 12:55
• @JustinBenfield That works for the open interval $(0,1)$, not the closed interval $[0,1]$. Commented May 9, 2016 at 12:56

No: $[0,1]$ is compact, $\mathbb R$ is not compact, and the image of a compact set under a continuous function is always compact.
The extreme value theorem, from real analysis, states that any continuous function from a compact set to the reals is bounded (and has a maximum). A bounded function cannot be surjective onto $\Bbb R$.