Proof required for following Theorem:

Let $X$ and $Y$ to be two spaces, each with two topologies, $\tau_1 , \tau_2$ and $\sigma_1 , \sigma_2 $ respectively. Suppose that $f : (X,\tau_1) \rightarrow (Y,\sigma_1)$ is a map from $X$ to $Y$ that is continuous with respect to the indicated topologies. Based on this prove that the following two statements are true:

(i) the title,

(ii) If $\sigma_2$ is coarser than $\sigma_1$, then $f : (X,\tau_1)\rightarrow (Y,\sigma_2)$ is continuous.

Definition of finer/coarser topologies:

Let $A$ be a set. Let $\tau$ and $\sigma$ be two topologies on $A$. Then $\tau$ is finer than $\sigma$ - and that $\sigma$ is coarser than $\tau$ - if $\sigma \subseteq \tau$.

General definition of continuity:

A function $f: X \rightarrow Y$ is continuous at $x\in X$ if for each neighbourhood $W$ of $f(x)$ there exists a neightbourhood $V$ of $x$ such that $f(V) \subseteq W$. We say that $f$ is continuous on $X$ if it is continuous at every $x \in X$.


The map $f : (X,\tau_1) \rightarrow (Y,\sigma_1)$ is continuous $\iff$ $\forall x\in X$ and each neighbourhood $W$ of $f(x)$, the pre-image of $f^{-1}(W)$ is a neighbourhood of $x$.

How do I use this information to formulate the proof?

  • 1
    $\begingroup$ Welcome to math.SE. Please tell us your own thoughts/work on the problem. That way we can offer help specifically addressed to your problems, if you don't the question will probably be closed for lack of context. $\endgroup$ – Henrik supports the community Aug 11 '18 at 9:49
  • $\begingroup$ The definitions are universal, so the reference is irrelevant. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Aug 11 '18 at 9:54
  • $\begingroup$ I have added some work. $\endgroup$ – L. Jetta Aug 11 '18 at 9:56

You better handle this leaving out elements of $X$ and their neighborhoods.

$f:(X,\tau_1)\to (Y,\sigma_1)$ continuous means exactly that: $$\{f^{-1}(U)\mid U\in\sigma_1)\}\subseteq\tau_1\tag1$$

$\tau_2$ finer than $\tau_1$ means exactly that $\tau_1\subseteq\tau_2$ and if that is the case we find on base of $(1)$ that:$$\{f^{-1}(U)\mid U\in\sigma_1)\}\subseteq\tau_2$$which means exactly that $f:(X,\tau_2)\to (Y,\sigma_1)$ is continuous.

$\sigma_2$ coarser than $\sigma_1$ means exactly that $\sigma_2\subseteq\sigma_1$ and if that is the case we find on base of $(1)$ that:$$\{f^{-1}(U)\mid U\in\sigma_2)\}\subseteq\tau_1$$which means exactly that $f:(X,\tau_1)\to (Y,\sigma_2)$ is continuous.

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  • $\begingroup$ Neat and clear! $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Aug 11 '18 at 10:11
  • $\begingroup$ Hmm, I wanted to approach it the way the earlier referenced book did but this will make do. $\endgroup$ – L. Jetta Aug 11 '18 at 10:13
  • $\begingroup$ @L.Jetta you can use the answer to study the theorem in terms of neighbourhoods. $\endgroup$ – Dog_69 Aug 11 '18 at 10:32

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