2

f:$\mathbb{R} \rightarrow \mathbb{R}$ such that f(x)=a, for all x$\in\mathbb{R}$ is continuous function.

$\mathbb{Q}$ is a subspace of $\mathbb{R}$ and hence f:$\mathbb{Q} \rightarrow \mathbb{R}$ such that f(x)=a, for all x$\in\mathbb{Q}$ is also continuous function with respect to subspace topology.

Using sequential continuity i can see that the function f:$\mathbb{Q} \rightarrow \mathbb{R}$ such that f(x)=a, for all x$\in\mathbb{Q}$ is continuous.

But if i will think in this way , {a} is closed in $\mathbb{R}$ and under continuous function inverse image of closed set is closed that is $f^{-1}$(a)=$\mathbb{Q}$ is closed. which is not possible because $\mathbb{Q}$ is not closed subspace of $\mathbb{R}$. hence f:$\mathbb{Q} \rightarrow \mathbb{R}$ such that f(x)=a, for all x$\in\mathbb{Q}$ is not continuous. where i am going wrong.

dipali mali
  • 555

1 Answers1

5

However, $\Bbb Q$ is closed in $\Bbb Q$.

When analysing the continuity of a function $g:A\to B$, then you need to analyse, for any open / closed $U\subseteq B$, whether $f^{-1}(U)$ is open / closed in $A$. It doesn't matter whether $A$ is thought of as a subspace of some larger space; in this context that larger space does not exist for our intents and purposes.

So in our case, when we have a function $f:\Bbb Q\to \Bbb R$, you must forget that $\Bbb Q$ is a part of $\Bbb R$, and just look at $\Bbb Q$ as a topological space in its own right.

Arthur
  • 199,419