Confusion in Subsets and subspace

Question

I know By, Heine Borel theorem that, subset $K$ of $\mathbb{R}$ is compact if and only if $K$ is closed and Bounded!

Now, if I consider $K$ as subspace of $\mathbb{R}$ then is

"a subspace $K$ of $\mathbb{R}$ is compact if and only if K is closed and Bounded subset of $\mathbb{R}$?

Further I think their huge difference in subset and subspace! for example, $[0,1]$ as subset of $\mathbb{R}$ is only closed whereas when we considered It as subspace, it is both open and closed

So my question is, is below statement is true?

a subspace $K$ of $\mathbb{R}$ is compact if and only if K is closed and Bounded subset of $\mathbb{R}$

Please I need helps, as I am beginner in "topology" (Edit: I don't know, reason of downvoting the question, as a beginner in topology one can have this question!)

Answer 1

Yes, the statement is true.

More generally we have:

Suppose $(X,\tau)$ is a topological space and $A\subset X$ is a subset. Let "$\tau_A$" denote the subspace topology on $A$ (precisely: $\tau_A=\{U\cap A: U\in\tau\}$). Then the following are equivalent: $(i)$ $A$ is a compact subset of $X$ in the sense of $\tau$. $(ii)$ The space $(A,\tau_A)$ is a compact space.

That is, this is a situation in which we can conflate between "set" and "space." As time goes on, you'll get a better sense of when we're in such a situation.

---

So why is the statement above true?

Let me sketch the proof in each direction.

$(ii)\implies (i)$: Suppose that $\{U_i:i\in I\}$ is an open cover (in the sense of $(X,\tau)$) of the set $A$, with no finite subcover. Let $V_i=U_i\cap A$. Then $\{V_i:i\in I\}$ is an open cover of $A$ with no finite subcover in the sense of $(A,\tau_A)$; that is, $(A,\tau_A)$ is not a compact space, and we can see that by "pushing down" a witness to the non-compactness of $A$ as a subset in $(X,\tau)$.

$(i)\implies (ii)$: This is the interesting one; we go the other way. Suppose $\{V_i:i\in I\}$ is an open cover of $A$ in the sense of $(A,\tau_A)$ with no finite subcover. Then by definition of the subspace topology, we can find open sets (in the sense of $\tau$) $U_i$ such that $U_i\cap A=V_i$. The set $\{U_i: i\in I\}$ is then an open cover of $A$ in the sense of $(X,\tau)$ with no finite subcover, so $A$ is not a compact subset of $(X,\tau)$.

Note that in the proof of $(i)\implies (ii)$ we implicitly used the axiom of choice, in picking a $U_i$ corresponding to each $V_i$. This simplified things a bit, but was unnecessary: we can explicitly define $$U_i=\bigcup_{W\in\tau, W\cap A=V_i} W,$$ and this doesn't need choice. (Note that here $U_i$ is the "biggest possible" choice of open set corresponding to $V_i$.)