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.


By definition of the inclusion map:

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

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


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.

Thanks in advance.

  • $\begingroup$ why $(a) i$ and not $i(a)$? $\endgroup$
    – user259242
    Sep 14, 2016 at 12:42
  • $\begingroup$ They're the same thing. $\endgroup$ Sep 14, 2016 at 12:42
  • $\begingroup$ 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. $\endgroup$
    – user259242
    Sep 14, 2016 at 12:44
  • $\begingroup$ @user259242 My apologies. Habits from Algebra die hard.. $\endgroup$ Sep 14, 2016 at 12:46
  • $\begingroup$ 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. $\endgroup$
    – user87690
    Sep 14, 2016 at 12:50

2 Answers 2


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.


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