# A Theorem Regarding Maps, the Trivial Topology and the Discrete Topology

My professor says these proofs are no good. Could you tell me what is wrong with them?

Theorem 1.a. If $Y$ has the trivial topology, then $f : X \rightarrow Y$ is continuous.

Proof. No matter what topology $X$ is equipped with, the sets $X$ and $\varnothing$ are open sets. The inverse image of the set $Y$ is $X$, which is an open set, and the inverse image of the set $\varnothing$ is $\varnothing$, which is an open set. Therefore $f$ is always continuous.

Theorem 1.b. If $X$ has the discrete topology, then $f : X \rightarrow Y$ is continuous.

Proof. If $X$ has the discrete topology then every subset of $X$ is an open set. No matter what topology $Y$ has, the inverse image of $Y$ is going to be a subset of $X$. Therefore the inverse image of $Y$ is always an open set. Therefore $f$ is always continuous.

• Proof of 1b is incorrect. You have to reason with any open subset of $Y$. – baharampuri Feb 5 '17 at 6:42
• Continuity requires that the inverse image of every open set in $Y$ is open in $X$. – Heath Winning Feb 5 '17 at 6:42
• @HeathWinning what are you talking about? By definition, $f^{-1}(Y)=\{x\in X:f(x)\in Y\}$, which is all of $X$ because $f$ is a function.... – Alex Mathers Feb 5 '17 at 6:44
• @HeathWinning If $f : X\to Y$ is a function, $f^{-1}(Y) = X$ always (every $x\in X$ must map to some $y\in Y$). As for 1b, I think what you meant to say but didn't quite say was that the inverse image of any subset of $Y$ is a subset of $X$. In particular, if $U\subseteq Y$ is open, then $f^{-1}(U)$ is open (as it is some subset of $X$). – Stahl Feb 5 '17 at 6:44
• Of course, my mistake. – Heath Winning Feb 5 '17 at 6:47

## 2 Answers

Your proofs are fine (the idea is good).

Maybe he/she wants a more pedantic/formal proof:

Suppose $f:X \rightarrow Y$ is a function such that $X$ is a space and $Y$ is indiscrete. Then $f$ is continuous.

Proof: let $O$ be open in $Y$. By the definition of the indiscrete topology, $O = Y$ or $O = \emptyset$. Then $f^{-1}[Y] = X$ is open in $X$ and $f^{-1}[\emptyset] = \emptyset$ is open in $X$ too. So $f^{-1}[O]$ is open in $X$ for all open subsets $O$ of $Y$. So $f$ is continuous.

Suppose $f:X \rightarrow Y$ is a function such that $Y$ is a space and $X$ is discrete. Then $f$ is continuous.

Proof: let $O$ be open in $Y$. Then $f^{-1}[O] \subset X$ and so as $\mathcal{T}_X = \mathscr{P}(X)$ (definition of a discrete space), $f^{-1}[O]$ is open in $X$ for any $O$ open in $Y$, so $f$ is continuous.

This spells out all the relevant definitions, so your professor knows you know them too.

• I think the professor's problem was just that OP wrote "$f^{-1}(Y)$ is open because $X$ is discrete", not "$f^{-1}(U)$ is open for all open $U\subset Y$ because $X$ is discrete". – Alex Mathers Feb 5 '17 at 6:46

I'm going to try to take a guess at what your professor disapproves of, because there are only very small issues with these proofs.

In $1.a$, you write that $f^{-1}(\{\})$ are $f^{-1}(Y)$ are open sets, but you don't mention that $\{\}$ and $Y$ are the only open sets of $Y$. This is the only thing wrong with this proof as far as I can tell.

In $1.b$, you mention that $f^{-1}(Y)$ is open, and your reasoning is correct, but this isn't every open subset of $Y$. Your reason actually shows that $f^{-1}(U)$ is open in $X$ for all open $U\subseteq Y$.

These seem like very minor details, but it's good that your professor wants to make sure you're learning to be very precise. You figured out the main points of the proofs on your own, so don't be discouraged. Just pay more attention to the finer details in the future and you'll be all set.