If $f$ is a continuous function from $\Bbb R$ to $\Bbb R^2$ and $A = (1,5]$. Is $f(A)$ connected?

Question

If $f$ is a continuous function from $\Bbb R$ to $\Bbb R^2$ and $A = (1,5]$. Is $f(A)$ connected?

Justify your answer...

Is this question correct? If yes kindly help me understand it. Thank you.

Answer 1

The most general result, as far as I know, is the following:

The image of a connected space under a continuous map is connected. More precisely, we can state the following: Let $X$ and $Y$ be any topological spaces, let $f \colon X \to Y$ be a continuous function, and let $A$ be a connected subspace of $X$. Then $f(A)$ is a connected subspace of $Y$.

For a proof, you can refer to Theorem 23.5 in the book Topology by James R. Munkres, 2nd edition.

And, we also have the following result:

Every interval --- open or closed and bounded or unbounded --- in $\mathbb{R}$ is connected.

For a proof, refer to Theorem 24.1 and Corollary 24.2 in Munkres.

Should you find any difficulty with either or both of the proofs mentioned above, please feel free to contact.

Hope this helps!

Answer 2

Yes, since $A=(1,5]$ is connected. In general, is you have a connected set $E\subset X$ and $f$ is a continuous function from a metric space $X$ to a metric space $Y$, then $f(E)$ is a connected set.

Answer 3

Assume $f(A)$ is disconnected. Then there exist open sets $U, V$ such that $U \cup V = f(A)$ and $U \cap V = \emptyset$. Then $f^{-1}(U) \cup f^{-1}(V) = A$, $f^{-1}(U) \cap f^{-1}(V) = \emptyset$ and both $f^{-1}(U)$ and $f^{-1}(V)$ are open by continuity of $f$. Therefore, $A$ is disconnected which is a contradiction since $A = (1, 5]$.