Continuity of inclusion map between subspace and topological space

Question:

Let $$A=\left [ 0,1 \right )$$ and $$X=\left [ 0,1 \right ]$$. Prove that if $$\left ( X,\alpha \right )$$ is a topological space and if A is given a subspace topology then the inclusion map i is continuous.

Attempt:

By definition of the inclusion map:

i:$$\left ( A,\beta \right )\rightarrow \left ( X,\alpha \right )$$.

$$a \mapsto \left ( a \right )i=a$$

Recall:

A function $$f:\left ( A,\beta \right )\rightarrow \left ( X,\alpha \right )$$ is called continuous IFF for every $$a \in \beta$$ we have $$f^{-1}\left ( a \right ) \in \alpha$$

It suffice to show that the pre-image of every element in $$\beta$$ is in $$\alpha$$.

The subspace topology $$\beta$$ on A is a topology so the empty set $$\varnothing \in \beta$$. Hence, by definition of the inclusion map, $$\left ( \varnothing \right )i^{-1}=\varnothing \in \alpha$$

$$A=\left [ 0,1 \right ) \in \beta$$.

$$\left ( A \right )i^{-1}=A \in \alpha$$ since $$A \subseteq X$$ and $$X \in \alpha$$ by definition of topology $$\alpha$$ on X.

I would like to ask if I am on the right track?

If I am not, any hints are appreciated.

• why $(a) i$ and not $i(a)$? Sep 14, 2016 at 12:42
• They're the same thing. Sep 14, 2016 at 12:42
• Its usually written with the function first, then the element. Writing it the other way is likely to confuse the people trying to help you. Sep 14, 2016 at 12:44
• @user259242 My apologies. Habits from Algebra die hard.. Sep 14, 2016 at 12:46
• You don't have to test particular sets like $∅$ or $A$. You can directly take any open $U ⊆ X$, consider what $i^{-1}(U)$ is, and observe that it's open in the subspace topology. Sep 14, 2016 at 12:50

You just have to keep in mind what the subspace topology is: a subset of $A$ belongs to $\beta$ if it is the intersection of $A$ with an element of $\alpha$.
But the preimage of a set under $\iota$ is exactly its intersection with $A$. So in particular the preimage of a set in $\alpha$ is the intersection of that set with $A$, which by definition of the subspace topology $\beta$ means that it belongs to $\beta$.
So preimages of sets in $\alpha$ belong to $\beta$, which means continuity.
Recall that a subset $U$ of $A$ is open in the subspace topology inherited from $X$ if and only if there exists a subset $V \in \alpha$ such that $U = A\cap\alpha$.
You're on the right track. Let $W$ be open in $X$. We need to show that $i^{-1}(W)$ is open in $A$. In other words, we need to find a subset $W'$ of $X$ which is open in $X$ and satisfies $i^{-1}(W) = W'\cap A$.
Now, $i^{-1}(W) = W\cap A$. But $W$ is open in $X$, so we're done.