Prove/Disprove that $X$ is compact

Suppose that $f: X \to Y$ is an injective function, where $Y$ has some topology $T_Y$. Equip $X$ with the topology $T_X :=\{f^{-1}(U):U\in T_Y\}$. Suppose that $Y$ is Hausdorff and compact.

I was wondering whether we can infer that $X$ is compact. I found out that $X$ is not necessarily compact, but I noticed that if $f(X)$ is closed, then $X$ is compact. I'm not sure of my work especially in the second one, so I want some critique. Thanks.

$1)$ Consider $X = [0,1[$, $Y = [0,1]$ and $f: X\to Y$, $f(x) = x$. Let $T_Y$ be the topology on $Y$ inherited from the usual topology on $\Bbb R$, $\tau$ be the topology on $X$ inherited from the usual topology on $\Bbb R$ and let's show that $\tau = T_X$. Let $a$ denote the usual topology on $\Bbb R$.

If $O \in \tau$, then $O = O' \cap [0,1[$ where $O' \in a$. Then $O'':=O'\setminus\{1\} \in a$ and $O = O'' \cap [0,1] \in T_Y$, hence $f^{-1}(O) \in T_X$, but $f^{-1}(O) = O$, so $O \in T_X$.

If $O \in T_X$, $O = f^{-1}(U)$, where $U \in T_Y$, so $U = U' \cap [0,1]$ where $U'\in a$. Then $$f^{-1}(U) = f^{-1}(U) \cap X = f^{-1}(U) \cap f^{-1}([0,1[) \\ = f^{-1}(U \cap [0,1[) = f^{-1}(U\setminus \{1\}) = f^{-1}(U' \cap [0,1[) = U' \cap [0,1[$$

So $O \in \tau$ and this proves $\tau = T_X$.

So in this example, $Y$ is Hausdorff and compact, $f$ is injective, but $X$ is not compact.

$2)$ Suppose that $f(X)$ is closed. Note that since $f$ is injective, $X$ is Hausdorff. We have $f:X \to f(X)$ is bijective. Let $g$ denote its inverse. We have $f(X)$ is compact since $Y$ is compact and $f(X)$ is closed. Now we claim that $(X,T_X) \cong (f(X), T_Y(f(X)))$, where $T_Y(f(X))$ is the topology on $f(X)$ inherited from $T_Y$. This will imply that $X$ is compact.

Proof: let's show that $g:(f(X),T_Y(f(X))) \to (X,T_X)$ is continuous. If $V \in T_X$, say $V = f^{-1}(U)$ where $U\in T_Y$, then: $g^{-1}(V) = g^{-1}(V) \cap f(X) = U \cap f(X)$, so $g^{-1}(V) \in T_Y(f(X))$ and $g$ is continuous. Now since $(f(X),T_Y(f(X)))$ is compact and $(X,T_X)$ is Hausdorff, we get that $g$ is a homeomorphism and the conclusion follows.

It’s correct, but it could be shortened a bit. In your counterexample you could note directly that $T_X$ is the relative topology that $X$ inherits from $Y$, and since $Y$ has the relative topology that it inherits from $\Bbb R$, $T_X$ must be $\tau$. In your proof of the general result when $f[X]$ is closed you don’t actually have to note that $g$ is a homeomorphism, though there’s certainly no harm in doing so: continuous maps preserve compactness, so continuity of $g$ is sufficient.
In fact you could observe more generally that if $Z=f[X]$, then $f$ is a homeomorphism between $X$ and $Z$: it’s a continuous, open bijection. Since $Y$ is compact Hausdorff, $Z$ (and hence also $X$) is compact if and only if $Z$ is closed in $Y$.
• @François: Your’e welcome! Perhaps the simplest counterexample is to let $X=\left\{\frac1n:n\in\Bbb Z^+\right\}$ and $Y=\{0\}\cup X$. Give $Y$ it’s usual topology, and $X$ will get its usual (discrete) topology. Take $f(x)=x$ for $x\in X$, as you did. This strips the example to its essentials. – Brian M. Scott Nov 16 '16 at 19:08