# A Language Belonges to PSPACE

Let $A,B$ be two languages, for which we know:

• $A \in PSPACE$
• $A\le_LB$

Can we conclude from the above that $B \in PSPACE$ ?

I think the answer is no, however I don't know how to prove it. I guess I have to write a reduction that proves otherwise, but how?

Thank you!

• Hint: LOGSPACE$\subseteq$PSPACE, and a logspace reduction has enough power to decide membership for a language in LOGSPACE. – Shaull Jun 30 '18 at 8:47
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No. Note that given languages $A,B$ and a reduction $A\le_L B$, you get that $B$ is at least as hard as $A$, but that's about it.
Take $A = \{0\}$, then $A\in \text{PSPACE}$, and take $B= H_{TM}\cup\{0\}$.
Define $f$ to be the following reduction $A \le_L H_{TM}\cup\{0\}$ (we can assume that $1\notin H_{TM}$):
$$f(0) = 0$$ $$\forall x\in \Sigma^*\setminus\{0\} . f(x)=1$$.
Clearly $f$ works in logarithmic space, but $B$ is clearly not in $\text{PSPACE}$. In fact, it isn't even in $\text{R}$.